Signals & Systems — Oppenheim, Willsky (2nd Edition)

Chapters 1–10: Key Theory & Practice Problems

Chapter 1: Signals and Systems

Key Concepts

  • Continuous-Time (CT) Signals: Defined for every value of time $t$, denoted $x(t)$.
  • Discrete-Time (DT) Signals: Defined only at integer values of $n$, denoted $x[n]$.
  • Signal Energy and Power:
    $$E_\infty = \int_{-\infty}^{\infty} |x(t)|^2\,dt \quad\text{(CT)}, \qquad E_\infty = \sum_{n=-\infty}^{\infty} |x[n]|^2 \quad\text{(DT)}$$
    $$P_\infty = \lim_{T\to\infty} \frac{1}{2T}\int_{-T}^{T} |x(t)|^2\,dt \quad\text{(CT average power)}$$
  • Transformations of Independent Variable:
    • Time shift: $x(t - t_0)$ shifts right by $t_0$
    • Time reversal: $x(-t)$ reflects about $t = 0$
    • Time scaling: $x(at)$, compressed if $|a| > 1$, expanded if $|a| < 1$
  • Periodic Signals: $x(t) = x(t + T)$ for all $t$. Fundamental period $T_0$ is the smallest positive $T$.
  • Even and Odd Decomposition:
    $$\text{Ev}\{x(t)\} = \tfrac{1}{2}[x(t) + x(-t)], \qquad \text{Od}\{x(t)\} = \tfrac{1}{2}[x(t) - x(-t)]$$
  • Complex Exponential Signals:
    $$\text{CT: } x(t) = Ce^{st}, \quad s = \sigma + j\omega \qquad\qquad \text{DT: } x[n] = Cz^n, \quad z = re^{j\omega}$$
    CT complex exponentials with distinct frequencies are always distinct. DT complex exponentials are periodic in frequency with period $2\pi$.
  • Unit Impulse and Unit Step:
    $$\text{DT: } \delta[n] = \begin{cases}1, & n=0\\0, & n\ne 0\end{cases}, \quad u[n] = \begin{cases}1, & n\ge 0\\0, & n < 0\end{cases}, \quad \delta[n] = u[n] - u[n-1]$$
    $$\text{CT: } u(t) = \begin{cases}1, & t>0\\0, & t<0\end{cases}, \quad \delta(t) = \frac{du(t)}{dt}, \quad \int_{-\infty}^{\infty}\delta(t)\,dt = 1$$
  • System Properties:
    • Memoryless: Output depends only on current input
    • Invertible: Distinct inputs produce distinct outputs
    • Causal: Output depends only on present and past inputs
    • Stable (BIBO): Bounded input $\Rightarrow$ bounded output
    • Time-Invariant: Time shift in input $\Rightarrow$ same time shift in output
    • Linear: Superposition: $T\{ax_1 + bx_2\} = aT\{x_1\} + bT\{x_2\}$

Problems and Solutions

Tier 1 · Important
Problem 1.1: Signal EnergyEasyTier 1?✓

Compute the total energy of $x(t) = e^{-2t}u(t)$.

Solution
$$E = \int_0^{\infty} |e^{-2t}|^2\,dt = \int_0^{\infty} e^{-4t}\,dt = \left[-\frac{1}{4}e^{-4t}\right]_0^{\infty} = \frac{1}{4}$$
Explanation

General definition. Signal energy is the integral (or sum) of the squared magnitude:

$$E = \int_{-\infty}^{\infty} |x(t)|^2\,dt \quad\text{(CT)}, \qquad E = \sum_{n=-\infty}^{\infty} |x[n]|^2 \quad\text{(DT)}$$

The $|\cdot|^2$ handles complex signals; for real signals it reduces to $x^2$. Classification:

  • Energy signal: $0 < E < \infty$ (e.g., a pulse or decaying exponential).
  • Power signal: $E = \infty$ but average power $P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2 dt$ is finite.
  • A signal is one or the other (or neither), never both.

By Parseval's theorem, energy is preserved between time and frequency domains: $\int |x(t)|^2 dt = \frac{1}{2\pi}\int |X(\omega)|^2 d\omega$.

Apply to $x(t) = e^{-2t}u(t)$. The unit step $u(t)$ restricts the integral to $t \ge 0$, where $x(t) = e^{-2t}$:

$$E = \int_{-\infty}^{\infty} |x(t)|^2\,dt = \int_0^{\infty} \bigl(e^{-2t}\bigr)^2 dt = \int_0^{\infty} e^{-4t}\,dt$$
$$= \left[-\frac{1}{4}e^{-4t}\right]_0^{\infty} = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$$

$E = \tfrac{1}{4}$ J — finite, so $x(t)$ is an energy signal.

Problem 1.2: Time TransformationMediumTier 1?✓

Given $x(t)$ defined for $0 \le t \le 2$, sketch $x(2t - 3)$.

Solution

Step 1: Replace $t$ with $t-3$: shift right by 3 → $x(t-3)$, defined on $[3, 5]$.
Step 2: Replace $t$ with $2t$: compress by factor 2 → $x(2t-3)$, defined when $0 \le 2t-3 \le 2$, i.e., $\frac{3}{2} \le t \le \frac{5}{2}$.
The signal is compressed by 2 and shifted to $[1.5,\; 2.5]$.

Problem 1.3: Even-Odd DecompositionEasyTier 1?✓

Decompose $x(t) = e^{t}u(t)$ into even and odd parts.

Solution
$$x_e(t) = \tfrac{1}{2}\bigl[e^{t}u(t) + e^{-t}u(-t)\bigr]$$
$$x_o(t) = \tfrac{1}{2}\bigl[e^{t}u(t) - e^{-t}u(-t)\bigr]$$
Problem 1.4: Periodicity of DT SignalsMediumTier 1?✓

Determine if $x[n] = \cos\!\left(\frac{6\pi n}{7}\right)$ is periodic. If so, find the fundamental period.

Solution

$\omega_0 = \frac{6\pi}{7}$. For periodicity, we need $\omega_0 N = 2\pi m$:

$$\frac{6\pi}{7}\,N = 2\pi m \;\Longrightarrow\; N = \frac{7m}{3}$$

Smallest integer $N$: $m=3 \Rightarrow N=7$. The signal is periodic with fundamental period $N=7$.

Problem 1.5: Periodicity of CT SignalsEasyTier 1?✓

Is $x(t) = \cos(2t) + \sin(\pi t)$ periodic?

Solution

$T_1 = \frac{2\pi}{2} = \pi$, $\;T_2 = \frac{2\pi}{\pi} = 2$. Ratio $\frac{T_1}{T_2} = \frac{\pi}{2}$ is irrational.
Since the ratio of periods is irrational, the sum is not periodic.

Problem 1.7: Linearity TestEasyTier 1?✓

Determine if the system $y(t) = t\cdot x(t)$ is linear.

Solution

Let $x_3(t) = ax_1(t)+bx_2(t)$. Then:

$$y_3(t) = t\bigl[ax_1(t)+bx_2(t)\bigr] = a\,t\,x_1(t) + b\,t\,x_2(t) = ay_1(t) + by_2(t)$$

Superposition holds $\Rightarrow$ the system is linear.

Tier 2 · Residual
Problem 1.8: Time Invariance TestMediumTier 2?✓

Is the system $y[n] = n\cdot x[n]$ time-invariant?

Solution

Shift input: $x_1[n] = x[n-n_0] \Rightarrow y_1[n] = n\,x[n-n_0]$.
Shift output: $y[n-n_0] = (n-n_0)\,x[n-n_0]$.
Since $n\,x[n-n_0] \ne (n-n_0)\,x[n-n_0]$, the system is not time-invariant.

Problem 1.9: Causality and StabilityEasyTier 2?✓

Determine if $y(t) = x(t-2)$ is (a) causal, (b) stable.

Solution

(a) Output at time $t$ depends on $x(t-2)$ (past input). The system is causal.
(b) If $|x(t)| \le M$, then $|y(t)| = |x(t-2)| \le M$. The system is BIBO stable.

Problem 1.10: Signal PowerEasyTier 2?✓

Compute the average power of $x(t) = A\cos(\omega_0 t + \phi)$.

Solution
$$P = \frac{1}{T}\int_0^{T} A^2\cos^2(\omega_0 t+\phi)\,dt = \frac{A^2}{2}$$

The average power of a sinusoid is half the square of its amplitude.

Problem 1.15: System Properties CombinedMediumTier 2?✓

For $y(t) = \int_{-\infty}^{t} x(\tau)\,d\tau$, determine linearity, time-invariance, causality, and stability.

Solution

Linear: Yes — integration is linear.
Time-Invariant: Yes — shifting input by $t_0$ shifts output by $t_0$.
Causal: Yes — integrates from $-\infty$ to present time $t$.
Stable: No — $x(t)=u(t)$ (bounded) gives $y(t)=t\,u(t)$ (unbounded).

Chapter 2: Linear Time-Invariant Systems

Key Concepts

  • Convolution Sum (DT): Any DT signal: $x[n] = \sum_k x[k]\,\delta[n-k]$. For LTI systems:
    $$y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k]$$
  • Convolution Integral (CT):
    $$y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$$
  • Properties of Convolution:
    • Commutative: $x*h = h*x$
    • Distributive: $x*(h_1+h_2) = x*h_1 + x*h_2$
    • Associative: $x*(h_1*h_2) = (x*h_1)*h_2$
    • Identity: $x[n]*\delta[n] = x[n]$
  • LTI System Properties via $h(t)$ or $h[n]$:
    • Memoryless: $h[n] = K\delta[n]$
    • Causal: $h(t) = 0$ for $t < 0$
    • Stable: $\int_{-\infty}^{\infty}|h(t)|\,dt < \infty$  or  $\sum_{n}|h[n]| < \infty$
  • Linear Constant-Coefficient Differential/Difference Equations:
    $$\text{CT: } \sum_{k=0}^{N} a_k \frac{d^k y}{dt^k} = \sum_{k=0}^{M} b_k \frac{d^k x}{dt^k}$$
    $$\text{DT: } \sum_{k=0}^{N} a_k\, y[n-k] = \sum_{k=0}^{M} b_k\, x[n-k]$$
  • Unit Step Response: $s(t) = \int_{-\infty}^{t} h(\tau)\,d\tau$,  so  $h(t) = \dfrac{ds(t)}{dt}$

Problems and Solutions

Tier 1 · Important
Problem 2.1: DT ConvolutionMediumTier 1?✓

Compute $y[n] = x[n]*h[n]$ where $x[n] = \{1, 2, 1\}$ and $h[n] = \{1, 1, 1\}$ (both starting at $n=0$).

Solution
$$\begin{aligned} y[0] &= 1\cdot1 = 1\\ y[1] &= 1\cdot1 + 2\cdot1 = 3\\ y[2] &= 1\cdot1 + 2\cdot1 + 1\cdot1 = 4\\ y[3] &= 2\cdot1 + 1\cdot1 = 3\\ y[4] &= 1\cdot1 = 1 \end{aligned}$$

$y[n] = \{1, 3, 4, 3, 1\}$ for $n = 0,1,2,3,4$.

Problem 2.2: CT Convolution with ExponentialsEasyTier 1?✓

Compute $y(t) = e^{-t}u(t) * u(t)$.

Solution
$$y(t) = \int_0^{t} e^{-\tau}\,d\tau = \bigl[-e^{-\tau}\bigr]_0^{t} = 1 - e^{-t}, \quad t \ge 0$$
$$y(t) = (1 - e^{-t})\,u(t)$$
Problem 2.3: Convolution of Two ExponentialsMediumTier 1?✓

Compute $y(t) = e^{-at}u(t) * e^{-bt}u(t)$ for $a \ne b$.

Solution
$$y(t) = \int_0^{t} e^{-a\tau}\,e^{-b(t-\tau)}\,d\tau = e^{-bt}\int_0^{t} e^{(b-a)\tau}\,d\tau$$
$$\boxed{y(t) = \frac{1}{a-b}\bigl(e^{-bt} - e^{-at}\bigr)\,u(t)}$$
Problem 2.4: Stability from Impulse ResponseEasyTier 1?✓

Determine if the system with $h(t) = e^{2t}u(-t)$ is stable.

Solution
$$\int_{-\infty}^{\infty}|h(t)|\,dt = \int_{-\infty}^{0} e^{2t}\,dt = \frac{1}{2} < \infty$$

The integral converges $\Rightarrow$ stable.

Problem 2.5: Causality and Stability CheckEasyTier 1?✓

Is the system with $h[n] = \left(\frac{1}{3}\right)^n u[n] + \left(\frac{1}{2}\right)^n u[n]$ causal and stable?

Solution

Causal: $h[n] = 0$ for $n < 0$ (both terms have $u[n]$) $\Rightarrow$ yes.
Stable:

$$\sum|h[n]| = \sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n + \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n = \frac{3}{2} + 2 = \frac{7}{2} < \infty \;\Rightarrow\; \text{yes.}$$
Problem 2.7: Cascade LTI SystemsMediumTier 1?✓

Two LTI systems in cascade: $h_1(t) = e^{-t}u(t)$ and $h_2(t) = e^{-2t}u(t)$. Find the overall impulse response.

Solution
$$h(t) = h_1(t)*h_2(t) = \frac{1}{2-1}\bigl(e^{-t} - e^{-2t}\bigr)u(t) = \bigl(e^{-t} - e^{-2t}\bigr)u(t)$$
Tier 2 · Residual
Problem 2.8: Difference EquationMediumTier 2?✓

Find the impulse response of $y[n] - \tfrac{1}{2}y[n-1] = x[n]$, assuming causality.

Solution

For $x[n] = \delta[n]$ with initial rest:

$$y[0] = 1,\quad y[1] = \tfrac{1}{2},\quad y[n] = \left(\frac{1}{2}\right)^n,\; n \ge 0$$
$$\boxed{h[n] = \left(\frac{1}{2}\right)^n u[n]}$$
Problem 2.10: Step ResponseEasyTier 2?✓

Given $h(t) = e^{-3t}u(t)$, find the unit step response $s(t)$.

Solution
$$s(t) = \int_0^{t} e^{-3\tau}\,d\tau = \frac{1}{3}\bigl(1 - e^{-3t}\bigr)\,u(t)$$
Problem 2.12: CT Differential EquationEasyTier 2?✓

Find $h(t)$ for $\dfrac{dy}{dt} + 2y(t) = x(t)$, assuming causality.

Solution
$$\boxed{h(t) = e^{-2t}\,u(t)}$$
Problem 2.15: Convolution with Delayed Impulse TrainEasyTier 2?✓

Compute $y[n] = x[n] * \bigl(\delta[n] + \delta[n-2]\bigr)$ where $x[n] = a^n u[n]$.

Solution
$$y[n] = x[n] + x[n-2] = a^n u[n] + a^{n-2}u[n-2]$$

Chapter 3: Fourier Series Representation of Periodic Signals

Key Concepts

  • Eigenfunctions of LTI Systems: If $x(t) = e^{st}$, then $y(t) = H(s)\,e^{st}$. Complex exponentials are eigenfunctions of LTI systems.
  • CT Fourier Series: For a periodic signal with period $T$, $\omega_0 = 2\pi/T$:
    $$x(t) = \sum_{k=-\infty}^{\infty} a_k\, e^{jk\omega_0 t}$$
    $$a_k = \frac{1}{T}\int_{T} x(t)\, e^{-jk\omega_0 t}\,dt$$
  • DT Fourier Series: For period $N$, $\omega_0 = 2\pi/N$:
    $$x[n] = \sum_{k=\langle N\rangle} a_k\, e^{jk(2\pi/N)n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]\, e^{-jk(2\pi/N)n}$$
  • Convergence (Dirichlet Conditions): Absolutely integrable over one period; finite maxima/minima; finite discontinuities.
  • Parseval's Relation:
    $$\frac{1}{T}\int_T |x(t)|^2\,dt = \sum_{k=-\infty}^{\infty}|a_k|^2 \qquad\text{(CT)}$$
    $$\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2 \qquad\text{(DT)}$$
  • Key Properties: Linearity; time shift $\to$ phase change $a_k e^{-jk\omega_0 t_0}$; conjugate symmetry for real signals $a_k = a_{-k}^*$; multiplication $\to$ convolution of coefficients.
  • Frequency Response: For periodic input, output coefficients: $b_k = H(jk\omega_0)\,a_k$.

Problems and Solutions

Tier 1 · Important
Problem 3.1: CT Fourier Coefficients of Square WaveMediumTier 1?✓

Find the Fourier series coefficients of a square wave: $x(t)=1$ for $|t|

Solution
$$a_k = \frac{1}{T}\int_{-T_1}^{T_1} e^{-jk\omega_0 t}\,dt = \frac{2\sin(k\omega_0 T_1)}{k\omega_0 T}, \qquad a_0 = \frac{2T_1}{T}$$
Problem 3.2: Fourier Series of CosineEasyTier 1?✓

Find the Fourier series coefficients of $x(t) = \cos(3t)$.

Solution
$$\cos(3t) = \tfrac{1}{2}e^{j3t} + \tfrac{1}{2}e^{-j3t} \;\Longrightarrow\; a_1 = \tfrac{1}{2},\; a_{-1} = \tfrac{1}{2},\; \text{all others } 0$$
Problem 3.3: DT Fourier SeriesEasyTier 1?✓

Find the Fourier series of $x[n] = \cos(2\pi n/4)$ with period $N=4$.

Solution
$$\cos\!\left(\frac{\pi n}{2}\right) = \tfrac{1}{2}e^{j(\pi/2)n} + \tfrac{1}{2}e^{-j(\pi/2)n}$$

$a_1 = \frac{1}{2}$, $a_3 = a_{-1} = \frac{1}{2}$ (since period 4), $a_0 = a_2 = 0$.

Problem 3.4: Parseval's Theorem ApplicationMediumTier 1?✓

Given the square wave from 3.1 with $T_1 = T/4$, compute $\sum|a_k|^2$.

Solution
$$\sum_{k=-\infty}^{\infty}|a_k|^2 = \frac{1}{T}\int_T |x(t)|^2\,dt = \frac{1}{T}\cdot 2T_1 = \frac{2T_1}{T} = \frac{1}{2}$$
Problem 3.5: Time Shift EffectEasyTier 1?✓

If $x(t)$ has coefficients $a_k$, find the coefficients of $y(t) = x(t - t_0)$.

Solution
$$b_k = a_k\, e^{-jk\omega_0 t_0}$$

Magnitudes $|b_k| = |a_k|$ are unchanged; only phase shifts.

Problem 3.7: RC Lowpass Filter on Square WaveMediumTier 1?✓

For $H(j\omega) = \frac{1}{1+j\omega RC}$, find the output Fourier coefficients for a square wave input with $a_k = \frac{\sin(k\pi/2)}{k\pi}$.

Solution
$$b_k = H(jk\omega_0)\,a_k = \frac{1}{1+jk\omega_0 RC}\cdot\frac{\sin(k\pi/2)}{k\pi}$$

Higher harmonics are attenuated by the lowpass filter.

Tier 2 · Residual
Problem 3.8: DT Impulse TrainMediumTier 2?✓

Find the Fourier series of $x[n] = \sum_r \delta[n-rN]$ (periodic impulse train, period $N$).

Solution
$$a_k = \frac{1}{N}\sum_{n=0}^{N-1}\delta[n]\,e^{-jk(2\pi/N)n} = \frac{1}{N} \quad \text{for all } k$$

All harmonics have equal amplitude $1/N$.

Problem 3.10: Sawtooth WaveHardTier 2?✓

Find the Fourier series of $x(t) = t/T$ for $-T/2 < t < T/2$.

Solution
$$a_0 = 0 \quad\text{(odd function)}$$
$$a_k = \frac{j(-1)^{k+1}}{2\pi k}, \quad k \ne 0$$
Problem 3.11: DT Rectangular PulseHardTier 2?✓

Find the DT Fourier series for $x[n]$ periodic with $N=5$: $x[n]=1$ for $n=0,1,2$ and $x[n]=0$ for $n=3,4$.

Solution
$$a_k = \frac{1}{5}\sum_{n=0}^{2} e^{-jk(2\pi/5)n} = \frac{1}{5}\cdot\frac{\sin(3\pi k/5)}{\sin(\pi k/5)}\,e^{-j2\pi k/5}, \qquad a_0 = \frac{3}{5}$$
Problem 3.12: Frequency Shaping (Differentiator)EasyTier 2?✓

A system has $H(j\omega) = j\omega$. If input has coefficients $a_k$, find output coefficients.

Solution
$$b_k = H(jk\omega_0)\,a_k = jk\omega_0\, a_k$$

This differentiator amplifies higher harmonics linearly with frequency.

Chapter 4: The Continuous-Time Fourier Transform

Key Concepts

  • Fourier Transform Pair:
    $$X(j\omega) = \int_{-\infty}^{\infty} x(t)\, e^{-j\omega t}\,dt \qquad\text{(Analysis)}$$
    $$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j\omega)\, e^{j\omega t}\,d\omega \qquad\text{(Synthesis)}$$
  • FT of Periodic Signals:
    $$x(t) = \sum_k a_k e^{jk\omega_0 t} \;\Longrightarrow\; X(j\omega) = 2\pi\sum_k a_k\,\delta(\omega - k\omega_0)$$
  • Key Transform Pairs:
    $$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j\omega},\; a>0 \qquad \delta(t) \longleftrightarrow 1 \qquad 1 \longleftrightarrow 2\pi\delta(\omega)$$
  • Properties:
    • Time shift: $x(t-t_0) \longleftrightarrow e^{-j\omega t_0}X(j\omega)$
    • Frequency shift: $e^{j\omega_0 t}x(t) \longleftrightarrow X(j(\omega-\omega_0))$
    • Differentiation: $\frac{dx}{dt} \longleftrightarrow j\omega\, X(j\omega)$
    • Convolution: $x*h \longleftrightarrow X\cdot H$
    • Multiplication: $x\cdot h \longleftrightarrow \frac{1}{2\pi}X*H$
    • Scaling: $x(at) \longleftrightarrow \frac{1}{|a|}X\!\left(\frac{j\omega}{a}\right)$
    • Parseval: $\int|x(t)|^2 dt = \frac{1}{2\pi}\int|X(j\omega)|^2 d\omega$
    • Duality: $X(jt) \longleftrightarrow 2\pi\, x(-\omega)$

Problems and Solutions

Tier 1 · Important
Problem 4.1: FT of Decaying ExponentialEasyTier 1?✓

Find $X(j\omega)$ for $x(t) = e^{-3t}u(t)$.

Solution
$$X(j\omega) = \int_0^{\infty} e^{-3t}e^{-j\omega t}\,dt = \int_0^{\infty}e^{-(3+j\omega)t}\,dt = \frac{1}{3+j\omega}$$
$$|X(j\omega)| = \frac{1}{\sqrt{9+\omega^2}}, \qquad \angle X(j\omega) = -\arctan\!\left(\frac{\omega}{3}\right)$$
Problem 4.2: FT of Two-Sided ExponentialMediumTier 1?✓

Find $X(j\omega)$ for $x(t) = e^{-|t|}$.

Solution
$$X(j\omega) = \int_{-\infty}^{0} e^{t} e^{-j\omega t}\,dt + \int_0^{\infty} e^{-t}e^{-j\omega t}\,dt = \frac{1}{1-j\omega} + \frac{1}{1+j\omega} = \frac{2}{1+\omega^2}$$
Problem 4.3: Inverse FTEasyTier 1?✓

Find $x(t)$ given $X(j\omega) = 2\pi\delta(\omega) + \pi\delta(\omega-4) + \pi\delta(\omega+4)$.

Solution
$$x(t) = \frac{1}{2\pi}\bigl[2\pi + \pi e^{j4t} + \pi e^{-j4t}\bigr] = 1 + \cos(4t)$$
Problem 4.4: Convolution PropertyMediumTier 1?✓

Find $y(t) = e^{-2t}u(t) * e^{-3t}u(t)$ using Fourier transforms.

Solution
$$Y(j\omega) = \frac{1}{(2+j\omega)(3+j\omega)} = \frac{1}{2+j\omega} - \frac{1}{3+j\omega}$$
$$\boxed{y(t) = \bigl(e^{-2t} - e^{-3t}\bigr)\,u(t)}$$
Problem 4.6: Differentiation PropertyEasyTier 1?✓

Find the FT of $y(t) = \frac{dx}{dt}$ where $x(t) = e^{-t}u(t)$.

Solution
$$Y(j\omega) = j\omega\cdot X(j\omega) = \frac{j\omega}{1+j\omega}$$

Note: $y(t) = \delta(t) - e^{-t}u(t)$, consistent with $\mathcal{F}\{\delta\}=1$.

Problem 4.7: Parseval's TheoremMediumTier 1?✓

Compute $\int_{-\infty}^{\infty}|X(j\omega)|^2\,d\omega$ where $X(j\omega) = \frac{1}{1+j\omega}$.

Solution
$$\int_{-\infty}^{\infty}|x(t)|^2\,dt = \int_0^{\infty} e^{-2t}\,dt = \frac{1}{2}$$
$$\frac{1}{2\pi}\int|X(j\omega)|^2\,d\omega = \frac{1}{2} \;\Longrightarrow\; \boxed{\int|X(j\omega)|^2\,d\omega = \pi}$$
Tier 2 · Residual
Problem 4.9: Rectangular PulseEasyTier 2?✓

Find the FT of $x(t) = 1$ for $|t| \le T$, $0$ otherwise.

Solution
$$X(j\omega) = \int_{-T}^{T} e^{-j\omega t}\,dt = \frac{2\sin(\omega T)}{\omega}$$
Problem 4.10: Frequency Shift (Modulation)EasyTier 2?✓

Find the FT of $y(t) = x(t)\cos(\omega_c t)$ in terms of $X(j\omega)$.

Solution
$$Y(j\omega) = \frac{1}{2}X(j(\omega-\omega_c)) + \frac{1}{2}X(j(\omega+\omega_c))$$

Multiplying by $\cos$ shifts the spectrum to $\pm\omega_c$ (amplitude modulation).

Problem 4.11: System Output via FTMediumTier 2?✓

$H(j\omega) = \frac{1}{2+j\omega}$, input $x(t) = e^{-t}u(t)$. Find $y(t)$.

Solution
$$Y(j\omega) = \frac{1}{(1+j\omega)(2+j\omega)} = \frac{1}{1+j\omega} - \frac{1}{2+j\omega}$$
$$y(t) = \bigl(e^{-t} - e^{-2t}\bigr)\,u(t)$$
Problem 4.15: FT of Impulse TrainHardTier 2?✓

Find the FT of $x(t) = \sum_{n=-\infty}^{\infty}\delta(t - nT)$.

Solution
$$X(j\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta\!\left(\omega - \frac{2\pi k}{T}\right) = \omega_s\sum_{k=-\infty}^{\infty}\delta(\omega - k\omega_s)$$

The FT of an impulse train with period $T$ is an impulse train with period $\omega_s = 2\pi/T$.

Chapter 5: The Discrete-Time Fourier Transform

Key Concepts

  • DTFT Pair:
    $$X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]\, e^{-j\omega n} \qquad\text{(Analysis)}$$
    $$x[n] = \frac{1}{2\pi}\int_{2\pi} X(e^{j\omega})\, e^{j\omega n}\,d\omega \qquad\text{(Synthesis)}$$
  • Key property: $X(e^{j\omega})$ is always periodic in $\omega$ with period $2\pi$.
  • Common Pairs:
    $$a^n u[n] \;\longleftrightarrow\; \frac{1}{1-ae^{-j\omega}}, \quad |a|<1 \qquad\qquad \delta[n] \;\longleftrightarrow\; 1$$
  • Properties:
    • Time shift: $x[n-n_0] \longleftrightarrow e^{-j\omega n_0}X(e^{j\omega})$
    • Frequency shift: $e^{j\omega_0 n}x[n] \longleftrightarrow X(e^{j(\omega-\omega_0)})$
    • Convolution: $x*h \longleftrightarrow X\cdot H$
    • Differencing: $x[n]-x[n-1] \longleftrightarrow (1-e^{-j\omega})X(e^{j\omega})$
    • Time expansion: $x_{(k)}[n] \longleftrightarrow X(e^{jk\omega})$
    • Parseval: $\sum|x[n]|^2 = \frac{1}{2\pi}\int_{2\pi}|X(e^{j\omega})|^2\,d\omega$

Problems and Solutions

Tier 1 · Important
Problem 5.1: DTFT of Geometric SequenceEasyTier 1?✓

Find $X(e^{j\omega})$ for $x[n] = (0.5)^n u[n]$.

Solution
$$X(e^{j\omega}) = \sum_{n=0}^{\infty}(0.5\,e^{-j\omega})^n = \frac{1}{1 - 0.5\,e^{-j\omega}}$$
Problem 5.2: DTFT of Finite-Length SequenceMediumTier 1?✓

Find $X(e^{j\omega})$ for $x[n] = 1$ for $0 \le n \le 4$, else $0$.

Solution
$$X(e^{j\omega}) = \sum_{n=0}^{4} e^{-j\omega n} = \frac{1 - e^{-j5\omega}}{1 - e^{-j\omega}} = e^{-j2\omega}\,\frac{\sin(5\omega/2)}{\sin(\omega/2)}$$
Problem 5.3: Inverse DTFTMediumTier 1?✓

Find $x[n]$ if $X(e^{j\omega}) = 1$ for $|\omega| \le W$, $0$ for $W < |\omega| \le \pi$.

Solution
$$x[n] = \frac{1}{2\pi}\int_{-W}^{W} e^{j\omega n}\,d\omega = \frac{\sin(Wn)}{\pi n}$$

This is the inverse DTFT of an ideal lowpass filter.

Problem 5.4: Convolution via DTFTMediumTier 1?✓

Find $y[n] = (0.5)^n u[n] * (0.25)^n u[n]$.

Solution
$$Y(e^{j\omega}) = \frac{1}{(1-0.5e^{-j\omega})(1-0.25e^{-j\omega})} = \frac{2}{1-0.5e^{-j\omega}} - \frac{1}{1-0.25e^{-j\omega}}$$
$$\boxed{y[n] = \bigl[2(0.5)^n - (0.25)^n\bigr]\,u[n]}$$
Problem 5.5: Time Shift PropertyEasyTier 1?✓

If $X(e^{j\omega}) = \frac{1}{1-0.5e^{-j\omega}}$, find the DTFT of $x[n-3]$.

Solution
$$\mathcal{F}\{x[n-3]\} = e^{-j3\omega}\,X(e^{j\omega}) = \frac{e^{-j3\omega}}{1-0.5e^{-j\omega}}$$
Problem 5.6: Parseval's for DTEasyTier 1?✓

Find $\sum_{n=-\infty}^{\infty}|(0.8)^n u[n]|^2$.

Solution
$$\sum_{n=0}^{\infty}(0.64)^n = \frac{1}{1-0.64} = \frac{25}{9}$$
Tier 2 · Residual
Problem 5.8: Two-Sided ExponentialHardTier 2?✓

Find the DTFT of $x[n] = a^{|n|}$ for $|a|<1$.

Solution
$$X(e^{j\omega}) = \frac{1}{1-ae^{j\omega}} - 1 + \frac{1}{1-ae^{-j\omega}} = \frac{1-a^2}{1-2a\cos\omega+a^2}$$
Problem 5.9: Difference Equation Frequency ResponseMediumTier 2?✓

Find $H(e^{j\omega})$ for $y[n] - 0.5y[n-1] = x[n] + x[n-1]$.

Solution
$$H(e^{j\omega}) = \frac{1+e^{-j\omega}}{1-0.5e^{-j\omega}}$$

$|H(e^{j0})| = 2/0.5 = 4$, $|H(e^{j\pi})| = 0/1.5 = 0$ $\Rightarrow$ lowpass filter.

Problem 5.11: DTFT of Unit StepHardTier 2?✓

Find the DTFT of $u[n]$.

Solution
$$U(e^{j\omega}) = \frac{1}{1-e^{-j\omega}} + \pi\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$

Includes an impulse at $\omega=0$ since $u[n]$ is not absolutely summable.

Problem 5.12: Accumulator Frequency ResponseHardTier 2?✓

$y[n] = \sum_{k=-\infty}^{n} x[k]$. Find $H(e^{j\omega})$.

Solution
$$H(e^{j\omega}) = \frac{1}{1-e^{-j\omega}} + \pi\,\delta(\omega)$$

The DT analogue of integration.

Chapter 6: Time and Frequency Characterization of Signals and Systems

Key Concepts

  • Magnitude-Phase: $H(j\omega) = |H(j\omega)|\,e^{j\angle H(j\omega)}$. Output: $|Y|=|H||X|$, $\angle Y = \angle H + \angle X$.
  • Linear Phase: $H(j\omega) = |H(j\omega)|\,e^{-j\omega t_d}$ means pure delay $t_d$ with no distortion.
  • Group Delay:
    $$\tau(\omega) = -\frac{d\,\angle H(j\omega)}{d\omega}$$
  • Bode Plots: $20\log_{10}|H(j\omega)|$ in dB vs $\log\omega$.
    • First-order pole at $-a$: $-20$ dB/decade above $\omega=a$
    • First-order zero at $-a$: $+20$ dB/decade above $\omega=a$
  • Ideal Filters: Impulse response is sinc (non-causal). Sharp cutoff $\Leftrightarrow$ infinite time response.
  • Second-Order System:
    $$H(j\omega) = \frac{\omega_n^2}{\omega_n^2 - \omega^2 + 2j\zeta\omega_n\omega}$$

Problems and Solutions

Tier 1 · Important
Problem 6.1: Bode PlotMediumTier 1?✓

Sketch the Bode magnitude plot for $H(j\omega) = \dfrac{10(j\omega+1)}{(j\omega+10)(j\omega+100)}$.

Solution

DC gain: $H(0) = \frac{10}{1000} = 0.01 = -40$ dB.
Zero at $\omega=1$: $+20$ dB/dec. Pole at $\omega=10$: net flat. Pole at $\omega=100$: $-20$ dB/dec.
Flat at $-40$dB until $\omega=1$, rises to $-20$dB at $\omega=10$, flat to $\omega=100$, then $-20$dB/dec.

Problem 6.3: Ideal Lowpass Impulse ResponseMediumTier 1?✓

Find $h(t)$ for an ideal lowpass filter with cutoff $\omega_c$ and unit gain.

Solution
$$h(t) = \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c} e^{j\omega t}\,d\omega = \frac{\sin(\omega_c t)}{\pi t}$$

Non-causal ($h(t)\ne 0$ for $t<0$), so ideal filters are not realizable.

Problem 6.4: First-Order RC CutoffEasyTier 1?✓

RC filter: $R=1\,\text{k}\Omega$, $C=1\,\mu\text{F}$. Find the 3dB cutoff.

Solution
$$\omega_c = \frac{1}{RC} = \frac{1}{10^3\cdot 10^{-6}} = 1000\;\text{rad/s}, \qquad f_c = \frac{\omega_c}{2\pi} \approx 159.2\;\text{Hz}$$
Problem 6.5: Second-Order ResonanceMediumTier 1?✓

$\omega_n = 10$, $\zeta = 0.1$. Find resonant frequency and peak gain.

Solution
$$\omega_r = \omega_n\sqrt{1-2\zeta^2} = 10\sqrt{0.98} \approx 9.9\;\text{rad/s}$$
$$|H|_{\max} = \frac{1}{2\zeta\sqrt{1-\zeta^2}} \approx 5.025 = 14.0\;\text{dB}$$
Problem 6.6: Linear Phase = No DistortionEasyTier 1?✓

Show that $\angle H(j\omega) = -\alpha\omega$ with $|H|=K$ introduces no distortion.

Solution
$$H(j\omega) = Ke^{-j\alpha\omega} \;\Longrightarrow\; Y(j\omega) = Ke^{-j\alpha\omega}X(j\omega) \;\Longrightarrow\; y(t) = K\,x(t-\alpha)$$

Scaled, delayed copy of input — no distortion.

Problem 6.7: DT First-Order SystemEasyTier 1?✓

Classify $H(e^{j\omega}) = \frac{1}{1-0.9e^{-j\omega}}$.

Solution
$$|H(e^{j0})| = \frac{1}{0.1} = 10, \qquad |H(e^{j\pi})| = \frac{1}{1.9} \approx 0.53$$

Lowpass filter with gain concentrated near $\omega=0$.

Tier 2 · Residual
Problem 6.8: Allpass FilterMediumTier 2?✓

Show $|H(j\omega)| = 1$ for $H(j\omega) = \frac{j\omega - a}{j\omega + a}$, $a>0$.

Solution
$$|H(j\omega)|^2 = \frac{(j\omega-a)(-j\omega-a)}{(j\omega+a)(-j\omega+a)} = \frac{\omega^2+a^2}{\omega^2+a^2} = 1$$
Problem 6.10: Moving Average FilterMediumTier 2?✓

$y[n] = \frac{1}{M}\sum_{k=0}^{M-1} x[n-k]$ with $M=4$. Find $H(e^{j\omega})$.

Solution
$$H(e^{j\omega}) = \frac{1}{4}\sum_{k=0}^{3}e^{-jk\omega} = \frac{1}{4}\,e^{-j3\omega/2}\,\frac{\sin(2\omega)}{\sin(\omega/2)}$$

Lowpass filter with nulls at $\omega = \pi/2,\;\pi,\; 3\pi/2$.

Problem 6.13: Filter TradeoffsEasyTier 2?✓

Explain the tradeoff between transition bandwidth and time-domain ringing.

Solution

Sharper transition band $\Rightarrow$ longer impulse response $\Rightarrow$ more ringing. Shorter impulse response $\Rightarrow$ wider transition band. This is the time-frequency uncertainty principle.

Problem 6.15: Rise Time vs BandwidthMediumTier 2?✓

First-order system $H(j\omega) = \frac{1}{1+j\omega/\omega_c}$. Relate rise time to bandwidth.

Solution
$$s(t) = (1-e^{-\omega_c t})\,u(t), \qquad t_r \approx \frac{2.2}{\omega_c}$$
$$t_r \cdot \text{BW} \approx 2.2 = \text{const.} \quad\Rightarrow\quad \text{wider BW} = \text{faster rise}$$

Chapter 7: Sampling

Key Concepts

  • Sampling Theorem (Nyquist): A bandlimited signal with $X(j\omega)=0$ for $|\omega|>\omega_M$ is uniquely determined by its samples if:
    $$\omega_s = \frac{2\pi}{T} \ge 2\omega_M \qquad\text{(Nyquist rate)}$$
  • Impulse-Train Sampling:
    $$X_s(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty} X(j(\omega - k\omega_s))$$
    Spectrum replicates at multiples of $\omega_s$.
  • Reconstruction: Ideal lowpass with cutoff $\omega_s/2$, gain $T$:
    $$x(t) = \sum_{n=-\infty}^{\infty} x(nT)\,\frac{\sin\bigl(\omega_c(t-nT)\bigr)}{\omega_c(t-nT)/\pi}$$
  • Aliasing: If $\omega_s < 2\omega_M$, spectral copies overlap — irreversible distortion.
  • Zero-Order Hold:
    $$H_0(j\omega) = T\,e^{-j\omega T/2}\,\frac{\sin(\omega T/2)}{\omega T/2}$$
  • DT Processing of CT Signals: $H_{\text{eff}}(j\omega) = H(e^{j\omega T})$ for $|\omega|<\pi/T$.
  • Decimation/Interpolation: Downsampling by $M$, upsampling by $L$.

Problems and Solutions

Tier 1 · Important
Problem 7.1: Nyquist RateEasyTier 1?✓

Find the Nyquist rate for $x(t) = 3\cos(50\pi t) + 10\sin(300\pi t) - \cos(100\pi t)$.

Solution
$$\omega_M = 300\pi\;\text{rad/s} \quad(f_M = 150\;\text{Hz})$$
$$\boxed{\text{Nyquist rate} = 2f_M = 300\;\text{Hz}}$$
Problem 7.2: Sampling a SinusoidEasyTier 1?✓

$x(t) = \cos(2\pi\cdot 100\,t)$ sampled at $f_s = 250$ Hz. Reconstructed frequency?

Solution

$f = 100 < f_s/2 = 125$ Hz $\Rightarrow$ no aliasing. Reconstructed signal: $\cos(2\pi\cdot 100\,t)$.

Problem 7.3: AliasingEasyTier 1?✓

$x(t) = \cos(2\pi\cdot 300\,t)$ sampled at $f_s = 500$ Hz. What frequency appears?

Solution
$$f = 300 > f_s/2 = 250 \;\Rightarrow\; f_{\text{alias}} = |f - f_s| = 200\;\text{Hz}$$
Problem 7.4: Maximum Sampling IntervalEasyTier 1?✓

$X(j\omega) = 0$ for $|\omega| > 1000\pi$. Maximum $T$?

Solution
$$T \le \frac{\pi}{\omega_M} = \frac{\pi}{1000\pi} = 1\;\text{ms}, \qquad f_s \ge 1000\;\text{Hz}$$
Problem 7.5: Reconstruction FormulaMediumTier 1?✓

Samples: $x[0]=1,\;x[1]=0,\;x[2]=-1,\;x[3]=0$ with $T=1$, ideal lowpass at $\omega_c=\pi$.

Solution
$$x(t) = \sum_n x[n]\,\frac{\sin(\pi(t-n))}{\pi(t-n)} = \text{sinc}(\pi t) - \text{sinc}(\pi(t-2))$$
Problem 7.6: Zero-Order Hold EffectMediumTier 1?✓

Describe the frequency response of a ZOH and its effect.

Solution
$$H_0(j\omega) = T\,e^{-j\omega T/2}\,\operatorname{sinc}\!\left(\frac{\omega T}{2\pi}\right)$$

Sinc-shaped magnitude causes droop at higher frequencies; introduces delay $T/2$.

Tier 2 · Residual
Problem 7.7: DT-to-CT Frequency MappingMediumTier 2?✓

CT signal ($\omega_M=2000\pi$) sampled at $\omega_s=8000\pi$. What CT frequency does $|\omega|\le\pi/4$ correspond to?

Solution
$$\omega_{\text{CT}} = \frac{\omega_{\text{DT}}}{T} = \omega_{\text{DT}}\cdot\frac{f_s}{1} \;\Rightarrow\; \frac{\pi/4}{2\pi}\cdot 8000\pi = 1000\pi\;\text{rad/s}$$
Problem 7.8: Anti-Aliasing FilterEasyTier 2?✓

Signal up to 5 kHz, sampling at 8 kHz. Design anti-aliasing filter.

Solution

Nyquist: $f_s/2 = 4$ kHz $<$ 5 kHz $\Rightarrow$ aliasing. Need lowpass with cutoff $\le 4$ kHz to remove 4–5 kHz content before sampling.

Problem 7.9: DecimationHardTier 2?✓

$X(e^{j\omega}) = 0$ for $\pi/3 < |\omega| \le \pi$. Downsample by 3: $y[n] = x[3n]$.

Solution
$$Y(e^{j\omega}) = \frac{1}{3}\sum_{k=0}^{2} X\!\left(e^{j(\omega-2\pi k)/3}\right)$$

Since $X$ is bandlimited to $|\omega|<\pi/3$, no overlap: $Y(e^{j\omega}) = \frac{1}{3}X(e^{j\omega/3})$.

Problem 7.10: Interpolation by $L=2$MediumTier 2?✓

Describe DT interpolation by factor 2.

Solution

Insert one zero between samples, then lowpass filter with cutoff $\pi/2$ and gain 2:

$$X_e(e^{j\omega}) = X(e^{j2\omega}) \quad\xrightarrow{\;\text{LPF}\;}\quad Y(e^{j\omega})$$

Chapter 8: Communication Systems

Key Concepts

  • Amplitude Modulation:
    $$y(t) = x(t)\cos(\omega_c t) \;\Longrightarrow\; Y(j\omega) = \tfrac{1}{2}X(j(\omega-\omega_c)) + \tfrac{1}{2}X(j(\omega+\omega_c))$$
  • Synchronous Demodulation:
    $$y(t)\cos(\omega_c t) = \tfrac{1}{2}x(t) + \tfrac{1}{2}x(t)\cos(2\omega_c t) \;\xrightarrow{\;\text{LPF}\;}\; \tfrac{1}{2}x(t)$$
  • Asynchronous (Envelope) Detection: $y(t) = [A+x(t)]\cos(\omega_c t)$. Works when $A+x(t)>0$.
  • FDM: Signals modulated to different carriers share one channel.
  • SSB: Transmit one sideband only — saves half the bandwidth.
  • FM: $\omega_i(t) = \omega_c + k_f x(t)$. More bandwidth but better noise immunity than AM.
  • Carson's Rule: $\text{BW} \approx 2(\Delta f + f_m)$.
  • TDM: Signals interleaved in time.

Problems and Solutions

Tier 1 · Important
Problem 8.1: AM BandwidthEasyTier 1?✓

$x(t)$ has bandwidth $W=5$ kHz. Bandwidth of $y(t) = x(t)\cos(2\pi\cdot 100\text{k}\cdot t)$?

Solution
$$\text{BW} = 2W = 10\;\text{kHz}$$

Spectrum at $[95, 105]$ kHz $\cup$ $[-105, -95]$ kHz.

Problem 8.2: Synchronous DemodulationEasyTier 1?✓

$y(t) = x(t)\cos(100t)$ demodulated by $\cos(100t)$ then lowpass. Output if BW of $x$ is 10 rad/s?

Solution
$$w(t) = x(t)\cos^2(100t) = \tfrac{1}{2}x(t) + \tfrac{1}{2}x(t)\cos(200t) \;\xrightarrow{\;\text{LPF}\;}\; \frac{x(t)}{2}$$
Problem 8.3: Phase Error in DemodulationMediumTier 1?✓

Demodulating carrier has phase error $\phi$. Output?

Solution
$$x(t)\cos(\omega_c t)\cdot\cos(\omega_c t+\phi) \;\xrightarrow{\;\text{LPF}\;}\; \frac{1}{2}x(t)\cos\phi$$

Attenuation by $\cos\phi$. At $\phi=90°$, output is zero.

Problem 8.4: FDM DesignEasyTier 1?✓

Three 4 kHz signals, 1 kHz guard bands.

Solution
$$\begin{aligned} \text{Signal 1:}&\quad f_c = 4\;\text{kHz} \;\to\; [0,\;8]\;\text{kHz}\\ \text{Signal 2:}&\quad f_c = 13\;\text{kHz} \;\to\; [9,\;17]\;\text{kHz}\\ \text{Signal 3:}&\quad f_c = 22\;\text{kHz} \;\to\; [18,\;26]\;\text{kHz} \end{aligned}$$
Problem 8.5: SSB vs DSB-SCEasyTier 1?✓

Compare bandwidth for 5 kHz baseband.

Solution
$$\text{DSB-SC: } 2\times 5 = 10\;\text{kHz}, \qquad \text{SSB: } 5\;\text{kHz}$$
Problem 8.6: Envelope Detection ConditionEasyTier 1?✓

$y(t) = [A+x(t)]\cos(\omega_c t)$, $\max|x(t)|=1$. Minimum $A$?

Solution
$$A + x(t) \ge 0 \;\;\forall\, t \;\Longrightarrow\; A \ge \max|x(t)| = 1$$
Tier 2 · Residual
Problem 8.7: AM Power EfficiencyMediumTier 2?✓

Conventional AM with tone modulation at index $m$.

Solution
$$\eta = \frac{m^2}{2+m^2}$$

At $m=1$: $\eta = \frac{1}{3} = 33\%$. Most power is in the carrier.

Problem 8.9: Narrowband FMHardTier 2?✓

Show that for $\beta \ll 1$, FM bandwidth $\approx$ AM bandwidth.

Solution
$$y(t) = A\cos(\omega_c t + \beta\sin\omega_m t) \approx A\cos\omega_c t - A\beta\sin\omega_m t\,\sin\omega_c t$$

Bandwidth $= 2f_m$, same as AM.

Problem 8.10: Wideband FM (Carson's Rule)EasyTier 2?✓

$f_m = 15$ kHz, $\Delta f = 75$ kHz.

Solution
$$\beta = \frac{\Delta f}{f_m} = 5, \qquad \text{BW} \approx 2(\Delta f + f_m) = 2(75+15) = 180\;\text{kHz}$$
Problem 8.11: TDM DesignEasyTier 2?✓

4 signals, each 4 kHz bandwidth, TDM multiplexed.

Solution
$$\text{Pulse rate} = 4\times 8000 = 32{,}000\;\text{pulses/s}, \qquad \text{BW} \ge 16\;\text{kHz}$$

Chapter 9: The Laplace Transform

Key Concepts

  • Bilateral Laplace Transform:
    $$X(s) = \int_{-\infty}^{\infty} x(t)\,e^{-st}\,dt, \qquad s = \sigma + j\omega$$
    Fourier transform is the special case $X(j\omega) = X(s)\big|_{s=j\omega}$ when $j\omega$ axis is in ROC.
  • Region of Convergence (ROC):
    • Right-sided signal: ROC right of the rightmost pole
    • Left-sided signal: ROC left of the leftmost pole
    • Two-sided signal: strip between poles
    • Causal + stable $\Rightarrow$ all poles in LHP, ROC includes $j\omega$ axis
  • Common Pairs:
    $$e^{-at}u(t) \;\longleftrightarrow\; \frac{1}{s+a},\;\text{ROC: Re}\{s\}>-a$$
    $$-e^{-at}u(-t) \;\longleftrightarrow\; \frac{1}{s+a},\;\text{ROC: Re}\{s\}<-a$$
  • Properties: Time shift ($e^{-st_0}X(s)$), $s$-domain shift, differentiation ($sX(s)$), convolution ($X\cdot H$), initial/final value theorems.
  • Unilateral Transform: Integration from $0^-$ to $\infty$. For ODEs with initial conditions.

Problems and Solutions

Tier 1 · Important
Problem 9.1: Laplace TransformEasyTier 1?✓

Find $X(s)$ and ROC for $x(t) = e^{-2t}u(t) + e^{-3t}u(t)$.

Solution
$$X(s) = \frac{1}{s+2} + \frac{1}{s+3} = \frac{2s+5}{(s+2)(s+3)}, \qquad \text{ROC: Re}\{s\}>-2$$
Problem 9.3: Inverse Laplace TransformMediumTier 1?✓

$X(s) = \frac{2s+4}{(s+1)(s+3)}$, ROC: Re$\{s\}>-1$.

Solution
$$X(s) = \frac{A}{s+1} + \frac{B}{s+3}: \quad A = \left.\frac{2s+4}{s+3}\right|_{s=-1} = 1,\quad B = \left.\frac{2s+4}{s+1}\right|_{s=-3} = 1$$
$$\boxed{x(t) = \bigl(e^{-t} + e^{-3t}\bigr)\,u(t)}$$
Problem 9.4: Pole-Zero Plot and StabilityEasyTier 1?✓

$H(s) = \frac{s+1}{(s+2)(s+3)}$. Is the causal system stable?

Solution

Zero at $s=-1$. Poles at $s=-2, -3$ (both in LHP).

$$\text{All poles Re}\{s\}<0 \;\Longrightarrow\; \text{causal and stable}$$
Problem 9.5: ODE via Unilateral LaplaceMediumTier 1?✓

$\frac{dy}{dt}+3y = e^{-t}u(t)$, $y(0^-)=1$.

Solution
$$sY(s) - 1 + 3Y(s) = \frac{1}{s+1} \;\Longrightarrow\; Y(s) = \frac{s+2}{(s+1)(s+3)} = \frac{1/2}{s+1} + \frac{1/2}{s+3}$$
$$\boxed{y(t) = \tfrac{1}{2}e^{-t} + \tfrac{1}{2}e^{-3t}, \quad t\ge 0}$$
Problem 9.6: Convolution via LaplaceMediumTier 1?✓

$y(t) = te^{-t}u(t) * e^{-2t}u(t)$.

Solution
$$Y(s) = \frac{1}{(s+1)^2(s+2)} = \frac{1}{(s+1)^2} - \frac{1}{s+1} + \frac{1}{s+2}$$
$$y(t) = \bigl(te^{-t} - e^{-t} + e^{-2t}\bigr)\,u(t)$$
Problem 9.7: Initial and Final Value TheoremsEasyTier 1?✓

$X(s) = \frac{3s+7}{(s+1)(s+2)}$.

Solution
$$x(0^+) = \lim_{s\to\infty} sX(s) = \lim_{s\to\infty}\frac{3s^2+7s}{s^2+3s+2} = 3$$
$$x(\infty) = \lim_{s\to 0} sX(s) = \frac{0}{2} = 0$$
Tier 2 · Residual
Problem 9.8: Transfer Function from Diff EqEasyTier 2?✓

$\frac{d^2y}{dt^2} + 5\frac{dy}{dt} + 6y = \frac{dx}{dt} + x$.

Solution
$$H(s) = \frac{s+1}{s^2+5s+6} = \frac{s+1}{(s+2)(s+3)}$$
Problem 9.9: Stability from PolesEasyTier 2?✓

(a) $H(s) = \frac{1}{s^2+4}$, (b) $H(s) = \frac{1}{s^2+2s+5}$.

Solution

(a) Poles at $s = \pm 2j$ (imaginary axis) $\Rightarrow$ marginally stable.
(b) Poles at $s = -1\pm 2j$ (LHP) $\Rightarrow$ stable.

Problem 9.11: Feedback SystemMediumTier 2?✓

Forward path $G(s)=\frac{10}{s+1}$, feedback $H(s)=1$. Closed-loop TF?

Solution
$$T(s) = \frac{G(s)}{1+G(s)H(s)} = \frac{10/(s+1)}{1+10/(s+1)} = \frac{10}{s+11}$$

Feedback moves the pole from $s=-1$ to $s=-11$.

Problem 9.12: ROC and Three Inverse TransformsHardTier 2?✓

$X(s) = \frac{1}{(s+1)(s-2)}$. Find all possible $x(t)$.

Solution

Partial fractions: $X(s) = \frac{1}{3}\!\left[\frac{-1}{s+1}+\frac{1}{s-2}\right]$.

$$\begin{aligned} \text{ROC: Re}\{s\}>2 &:\quad x(t) = \tfrac{1}{3}(-e^{-t}+e^{2t})u(t) &&\text{(causal, unstable)}\\ \text{ROC: }-1<\text{Re}\{s\}<2 &:\quad x(t) = \tfrac{1}{3}(e^{-t}u(t)+e^{2t}u(-t)) &&\text{(stable)}\\ \text{ROC: Re}\{s\}<-1 &:\quad x(t) = \tfrac{1}{3}(e^{-t}u(-t)-e^{2t}u(-t)) &&\text{(anti-causal)} \end{aligned}$$

Chapter 10: The z-Transform

Key Concepts

  • z-Transform:
    $$X(z) = \sum_{n=-\infty}^{\infty} x[n]\,z^{-n}, \qquad z = re^{j\omega}$$
    DTFT: $X(e^{j\omega}) = X(z)\big|_{z=e^{j\omega}}$ when unit circle is in ROC.
  • ROC Rules:
    • Causal (right-sided): ROC outside outermost pole
    • Left-sided: ROC inside innermost pole
    • Finite-length: entire $z$-plane (possibly excluding $z=0$ or $\infty$)
    • Causal + stable: all poles inside unit circle
  • Common Pairs:
    $$a^n u[n] \;\longleftrightarrow\; \frac{1}{1-az^{-1}} = \frac{z}{z-a}, \quad\text{ROC: }|z|>|a|$$
    $$-a^n u[-n-1] \;\longleftrightarrow\; \frac{1}{1-az^{-1}}, \quad\text{ROC: }|z|<|a|$$
  • Properties: Time shift ($z^{-n_0}X(z)$), convolution ($X\cdot H$), $z$-domain differentiation ($-z\frac{dX}{dz} \longleftrightarrow n\,x[n]$).
  • System Analysis: $H(z)=Y(z)/X(z)$. Causal+stable $\Leftrightarrow$ all poles inside unit circle. FIR: always stable.
  • Mapping: $z = e^{sT}$ relates $z$-transform to Laplace through sampling.

Problems and Solutions

Tier 1 · Important
Problem 10.1: z-Transform of Causal ExponentialEasyTier 1?✓

$x[n] = (0.5)^n u[n]$.

Solution
$$X(z) = \sum_{n=0}^{\infty}(0.5z^{-1})^n = \frac{1}{1-0.5z^{-1}} = \frac{z}{z-0.5}, \qquad \text{ROC: }|z|>0.5$$
Problem 10.2: ROC IntersectionMediumTier 1?✓

$x[n] = (0.5)^n u[n] + (-\frac{1}{3})^n u[-n-1]$.

Solution

First term ROC: $|z|>0.5$. Second term (left-sided): ROC $|z|<\frac{1}{3}$.
Intersection is empty ($\frac{1}{3}<0.5$) $\Rightarrow$ z-transform does not exist for this signal.

Problem 10.3: Inverse z-TransformMediumTier 1?✓

$X(z) = \frac{z}{(z-0.5)(z-0.25)}$, ROC: $|z|>0.5$.

Solution
$$X(z) = \frac{2}{1-0.5z^{-1}} - \frac{1}{1-0.25z^{-1}}$$
$$\boxed{x[n] = \bigl[2(0.5)^n - (0.25)^n\bigr]\,u[n]}$$
Problem 10.4: System Function from Difference EquationMediumTier 1?✓

$y[n] - 0.5y[n-1] + 0.06y[n-2] = x[n]$.

Solution
$$H(z) = \frac{1}{1-0.5z^{-1}+0.06z^{-2}} = \frac{1}{(1-0.2z^{-1})(1-0.3z^{-1})}$$

Poles at $z=0.2, 0.3$ — both inside unit circle $\Rightarrow$ causal and stable.

Problem 10.5: FIR SystemMediumTier 1?✓

$y[n] = x[n] - x[n-2]$.

Solution
$$H(z) = 1 - z^{-2}, \qquad H(e^{j\omega}) = 1-e^{-j2\omega} = -2j\sin(\omega)\,e^{-j\omega}$$
$$|H(e^{j\omega})| = 2|\sin\omega| \quad\Rightarrow\quad \text{highpass (zero at } \omega=0\text{)}$$
Problem 10.6: Stability CheckEasyTier 1?✓

Causal system: $H(z) = \frac{1}{1-1.5z^{-1}}$. Stable?

Solution
$$\text{Pole at } z=1.5 \;(|z|>1) \;\Longrightarrow\; h[n] = (1.5)^n u[n] \to \infty$$

Not stable.

Tier 2 · Residual
Problem 10.7: Difference Equation with Initial ConditionsHardTier 2?✓

$y[n]-0.5y[n-1] = \delta[n]+\delta[n-1]$, $y[-1]=2$. Unilateral z-transform.

Solution
$$Y(z)(1-0.5z^{-1}) = 1+z^{-1}+0.5\cdot 2 = 2+z^{-1}$$
$$Y(z) = \frac{2+z^{-1}}{1-0.5z^{-1}} \;\Longrightarrow\; y[n] = 2(0.5)^n u[n] + (0.5)^{n-1}u[n-1]$$
Problem 10.11: Direct Form II RealizationMediumTier 2?✓

$H(z) = \frac{1+2z^{-1}}{1-0.5z^{-1}}$.

Solution
$$w[n] = 0.5\,w[n-1]+x[n], \qquad y[n] = w[n]+2\,w[n-1]$$

One delay element stores $w[n-1]$, used in both paths.

Problem 10.14: Frequency Response from Poles/ZerosMediumTier 2?✓

Zero at $z=1$, pole at $z=0.95$. Sketch $|H(e^{j\omega})|$.

Solution
$$|H(e^{j0})| = \frac{|e^{j0}-1|}{|e^{j0}-0.95|} = \frac{0}{0.05} = 0$$
$$|H(e^{j\pi})| = \frac{|e^{j\pi}-1|}{|e^{j\pi}-0.95|} = \frac{2}{1.95} \approx 1.03$$

Highpass filter.

Problem 10.15: $z = e^{sT}$ MappingMediumTier 2?✓

Show how $z = e^{sT}$ connects $z$-transform to Laplace.

Solution
$$X(z) = \sum_n x_c(nT)\,z^{-n} \;\xrightarrow{z=e^{sT}}\; X(e^{sT}) = \sum_n x_c(nT)\,e^{-snT}$$

Key mapping:

$$\begin{aligned} s\text{-plane } j\omega\text{ axis} &\;\longleftrightarrow\; z\text{-plane unit circle}\\ \text{LHP (Re}\{s\}<0) &\;\longleftrightarrow\; |z|<1\\ \text{RHP (Re}\{s\}>0) &\;\longleftrightarrow\; |z|>1 \end{aligned}$$

Stable CT (poles in LHP) $\Leftrightarrow$ stable DT (poles inside unit circle).

Others — Residual Problems

Auxiliary problems moved out of the main 10-problem-per-chapter set. Useful but secondary in priority.

Chapter 1: Signals and Systems

Problem 1.6: Unit Step Representation?✓

Express the rectangular pulse $p(t) = 1$ for $|t| \le 1$, $0$ otherwise, using unit step functions.

Solution
$$p(t) = u(t+1) - u(t-1)$$
Problem 1.11: DT Signal Energy?✓

Compute the energy of $x[n] = \left(\frac{1}{2}\right)^n u[n]$.

Solution
$$E = \sum_{n=0}^{\infty}\left|\left(\frac{1}{2}\right)^n\right|^2 = \sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n = \frac{1}{1-\frac{1}{4}} = \frac{4}{3}$$
Problem 1.12: Memoryless System?✓

Classify: (a) $y[n] = x[n]^2$, (b) $y[n] = x[n-1]$.

Solution

(a) Output depends only on current input $\Rightarrow$ memoryless.
(b) Output depends on past input $\Rightarrow$ has memory.

Problem 1.13: Invertibility?✓

Is the system $y(t) = x^2(t)$ invertible?

Solution

Both $x(t)$ and $-x(t)$ produce the same output $y(t) = x^2(t)$. Distinct inputs map to the same output $\Rightarrow$ not invertible.

Problem 1.14: DT Complex Exponentials?✓

Show that $e^{j\omega n} = e^{j(\omega+2\pi)n}$ for all integer $n$.

Solution
$$e^{j(\omega+2\pi)n} = e^{j\omega n}\cdot e^{j2\pi n} = e^{j\omega n}\cdot 1 = e^{j\omega n}$$

Since $n$ is an integer, $e^{j2\pi n}=1$. DT complex exponentials are periodic in frequency with period $2\pi$.

Chapter 2: Linear Time-Invariant Systems

Problem 2.6: Convolution with Impulse?✓

Compute $x(t) * \delta(t-3)$.

Solution
$$x(t) * \delta(t-3) = x(t-3)$$

Convolution with a shifted impulse simply shifts the signal.

Problem 2.9: Rectangular Pulse Convolution?✓

Compute the convolution of $p(t) = u(t) - u(t-1)$ with itself.

Solution
$$y(t) = p(t)*p(t) = \begin{cases} t, & 0 \le t \le 1\\ 2-t, & 1 \le t \le 2\\ 0, & \text{otherwise} \end{cases}$$

The result is a triangular pulse.

Problem 2.11: Parallel LTI Systems?✓

$h_1[n] = \left(\frac{1}{2}\right)^n u[n]$ and $h_2[n] = -\left(\frac{1}{3}\right)^n u[n]$ in parallel.

Solution
$$h[n] = h_1[n] + h_2[n] = \left(\frac{1}{2}\right)^n u[n] - \left(\frac{1}{3}\right)^n u[n]$$
Problem 2.13: Commutativity Verification?✓

Verify $x[n]*h[n] = h[n]*x[n]$ for $x[n] = \delta[n] + 2\delta[n-1]$ and $h[n] = u[n]-u[n-3]$.

Solution

$h[n] = \{1,1,1\}$ for $n=0,1,2$.

$$x*h:\; y = \{1,\, 3,\, 3,\, 2\} \qquad h*x:\; y = \{1,\, 3,\, 3,\, 2\}$$

Both identical $\Rightarrow$ commutativity verified.

Problem 2.14: Unstable System?✓

Show that the system with $h(t) = u(t)$ is not BIBO stable.

Solution
$$\int_{-\infty}^{\infty}|u(t)|\,dt = \int_0^{\infty} 1\,dt = \infty$$

Counterexample: $x(t)=u(t)$ gives $y(t)=t\,u(t) \to \infty$. Not stable.

Chapter 3: Fourier Series

Problem 3.6: Fourier Series from Symmetry?✓

Show that for a real, even periodic signal, $a_k$ is real and even.

Solution

Real signal: $a_k = a_{-k}^*$ (conjugate symmetry).
Even signal: $a_k = a_{-k}$.
Combining: $a_k = a_k^* \Rightarrow a_k$ is real. And $a_k = a_{-k} \Rightarrow$ even.

Problem 3.9: Product of Two Periodic Signals?✓

If $x(t)$ has coefficients $a_k$ and $y(t)$ has $b_k$ (same period), find $c_k$ for $z(t)=x(t)y(t)$.

Solution
$$c_k = \sum_{l=-\infty}^{\infty} a_l\, b_{k-l}$$

Multiplication in time $\longleftrightarrow$ convolution of coefficients.

Problem 3.13: Gibbs Phenomenon?✓

Explain the Gibbs phenomenon for the Fourier series of a square wave.

Solution

Truncating to $N$ terms causes oscillatory overshoot near discontinuities. As $N\to\infty$, the overshoot narrows but never vanishes, with peak overshoot $\approx 9\%$ of the discontinuity height.

Problem 3.14: First-Order DT Filter?✓

$y[n] = \frac{1}{2}y[n-1] + x[n]$ receives periodic input with period 4. Find output coefficients.

Solution
$$H(e^{j\omega}) = \frac{1}{1 - \frac{1}{2}e^{-j\omega}}, \qquad b_k = \frac{a_k}{1 - \frac{1}{2}e^{-jk\pi/2}}$$
Problem 3.15: Power Spectrum?✓

$a_0=1$, $a_1=a_{-1}=0.5$, $a_2=a_{-2}=0.2$, all others $0$. Find the average power.

Solution
$$P = |a_0|^2 + 2|a_1|^2 + 2|a_2|^2 = 1 + 2(0.25) + 2(0.04) = 1.58$$

DC power fraction: $1/1.58 = 63.3\%$. Fundamental: $0.5/1.58 = 31.6\%$.

Chapter 4: CT Fourier Transform

Problem 4.5: Time Scaling?✓

If $X(j\omega) = \frac{1}{1+j\omega}$, find the FT of $x(2t)$.

Solution
$$\mathcal{F}\{x(2t)\} = \frac{1}{2}X\!\left(\frac{j\omega}{2}\right) = \frac{1}{2}\cdot\frac{1}{1+j\omega/2} = \frac{1}{2+j\omega}$$
Problem 4.8: Duality?✓

Use duality to find the FT of $x(t) = \frac{1}{1+t^2}$.

Solution

Known: $e^{-|t|} \longleftrightarrow \frac{2}{1+\omega^2}$. By duality: $\frac{2}{1+t^2} \longleftrightarrow 2\pi\,e^{-|\omega|}$.

$$\boxed{\frac{1}{1+t^2} \;\longleftrightarrow\; \pi\,e^{-|\omega|}}$$
Problem 4.12: Integration Property?✓

Find the FT of $y(t) = \int_{-\infty}^{t} e^{-\tau}u(\tau)\,d\tau$.

Solution
$$Y(j\omega) = \frac{X(j\omega)}{j\omega} + \pi X(0)\,\delta(\omega) = \frac{1}{j\omega(1+j\omega)} + \pi\delta(\omega)$$

$y(t) = (1 - e^{-t})u(t)$.

Problem 4.13: Bandwidth and Time-Width?✓

Show that compressing $x(t) = e^{-|t|}$ in time expands its bandwidth.

Solution
$$X(j\omega) = \frac{2}{1+\omega^2} \quad\Rightarrow\quad \text{3dB BW at } \omega=1$$
$$\mathcal{F}\{x(2t)\} = \frac{1}{1+(\omega/2)^2} \quad\Rightarrow\quad \text{3dB BW at } \omega=2$$

Compressing time by 2 doubles the bandwidth — a fundamental time-frequency trade-off.

Problem 4.14: Conjugate Symmetry?✓

Show that for a real signal, $|X(j\omega)|$ is even and $\angle X(j\omega)$ is odd.

Solution

For real $x(t)$: $X(-j\omega) = X^*(j\omega)$. Therefore:

$$|X(-j\omega)| = |X^*(j\omega)| = |X(j\omega)| \quad\Rightarrow\quad \text{magnitude is even}$$
$$\angle X(-j\omega) = -\angle X(j\omega) \quad\Rightarrow\quad \text{phase is odd}$$

Chapter 5: DT Fourier Transform

Problem 5.7: Frequency Shift?✓

Find the DTFT of $y[n] = e^{j\pi n/4}\,x[n]$.

Solution
$$Y(e^{j\omega}) = X(e^{j(\omega-\pi/4)})$$
Problem 5.10: Multiplication by $(-1)^n$?✓

Find $Y(e^{j\omega})$ for $y[n] = x[n]\,(-1)^n$.

Solution
$$(-1)^n = e^{j\pi n} \;\Longrightarrow\; Y(e^{j\omega}) = X(e^{j(\omega-\pi)})$$

Shifts spectrum by $\pi$: converts lowpass $\leftrightarrow$ highpass.

Problem 5.13: DT Duality?✓

Find the DTFT of $x[n] = 1$ for all $n$.

Solution
$$X(e^{j\omega}) = 2\pi\sum_{k=-\infty}^{\infty}\delta(\omega - 2\pi k)$$
Problem 5.14: Time Expansion?✓

If $x[n] = (0.5)^n u[n]$ and $x_{(2)}[n]$ inserts zeros, find the DTFT.

Solution
$$X_{(2)}(e^{j\omega}) = X(e^{j2\omega}) = \frac{1}{1-0.5e^{-j2\omega}}$$
Problem 5.15: Frequency Differentiation?✓

Find the DTFT of $y[n] = n\,a^n u[n]$.

Solution

Using $n\,x[n] \longleftrightarrow j\,\frac{dX(e^{j\omega})}{d\omega}$:

$$Y(e^{j\omega}) = \frac{ae^{-j\omega}}{(1-ae^{-j\omega})^2}$$

Chapter 6: Time & Freq Characterization

Problem 6.2: Group Delay?✓

Find $\tau(\omega)$ for $H(j\omega) = \frac{e^{-j3\omega}}{1+j\omega}$.

Solution
$$\angle H = -3\omega - \arctan(\omega) \;\Longrightarrow\; \tau(\omega) = 3 + \frac{1}{1+\omega^2}$$
Problem 6.9: Cascaded Bandwidth?✓

Two identical first-order lowpass filters (cutoff $\omega_c$) cascaded. Find 3dB bandwidth.

Solution
$$\frac{1}{(1+(\omega/\omega_c)^2)^2} = \frac{1}{2} \;\Longrightarrow\; \omega_{3\text{dB}} = \omega_c\sqrt{\sqrt{2}-1} \approx 0.644\,\omega_c$$
Problem 6.11: DT Second-Order Bandpass?✓

Poles at $z = 0.9e^{\pm j\pi/4}$. Where is the peak response?

Solution

Pole angle $= \pi/4$, so $|H(e^{j\omega})|$ peaks near $\omega = \pi/4$. This is a bandpass filter.

Problem 6.12: Gain in dB?✓

Convert $|H| = 0.01,\; 1,\; 100$ to dB.

Solution
$$20\log_{10}(0.01) = -40\;\text{dB}, \quad 20\log_{10}(1) = 0\;\text{dB}, \quad 20\log_{10}(100) = 40\;\text{dB}$$
Problem 6.14: Highpass from Lowpass?✓

Design a DT highpass from a lowpass $H_{lp}(e^{j\omega})$.

Solution
$$H_{hp}(e^{j\omega}) = 1 - H_{lp}(e^{j\omega})$$

Or multiply impulse response by $(-1)^n$ to shift spectrum by $\pi$.

Chapter 7: Sampling

Problem 7.11: CT-to-DT Frequency Mapping?✓

$x(t) = \cos(6000\pi t)$ sampled at 8000 samples/sec.

Solution
$$\omega = \Omega T = \frac{2\pi f}{f_s} = \frac{2\pi\cdot 3000}{8000} = \frac{3\pi}{4} \;\Longrightarrow\; x[n] = \cos\!\left(\frac{3\pi n}{4}\right)$$
Problem 7.12: Oversampling Benefit?✓

Why is oversampling practically useful?

Solution

Wider transition band for anti-aliasing filter (simpler analog filter). Reduces in-band quantization noise via noise shaping. Easier reconstruction.

Problem 7.13: Digital Differentiator?✓

Frequency response of DT differentiator for CT signals sampled at $1/T$.

Solution
$$H_d(e^{j\omega}) = \frac{j\omega}{T}, \quad |\omega| \le \pi$$
Problem 7.14: Bandpass Sampling?✓

Bandpass signal: 900–1100 Hz. Minimum sampling rate?

Solution
$$B = 200\;\text{Hz},\; f_H = 1100,\; n = \left\lfloor\frac{f_H}{B}\right\rfloor = 5$$
$$f_s = \frac{2f_H}{n} = \frac{2200}{5} = 440\;\text{Hz}$$
Problem 7.15: ZOH Attenuation?✓

ZOH with $T=1$ ms. Attenuation at $f=400$ Hz?

Solution
$$\left|\frac{\sin(\pi f T)}{\pi f T}\right| = \frac{\sin(0.4\pi)}{0.4\pi} \approx 0.757 \;\Rightarrow\; 20\log_{10}(0.757) \approx -2.4\;\text{dB}$$

Chapter 8: Communication Systems

Problem 8.8: PAM Bandwidth?✓

4 kHz signal, PAM at 8 kHz. Minimum transmission bandwidth?

Solution
$$\text{BW} \ge \frac{f_s}{2} = 4\;\text{kHz}$$
Problem 8.12: Superheterodyne Receiver?✓

$f_c = 1$ MHz, $f_{\text{LO}} = 1.455$ MHz.

Solution
$$f_{\text{IF}} = f_{\text{LO}} - f_c = 455\;\text{kHz}, \qquad f_{\text{image}} = f_{\text{LO}} + f_{\text{IF}} = 1.91\;\text{MHz}$$
Problem 8.13: DT Modulation?✓

$y[n] = x[n]\cos(\pi n/4)$. Spectrum?

Solution
$$Y(e^{j\omega}) = \tfrac{1}{2}X(e^{j(\omega-\pi/4)}) + \tfrac{1}{2}X(e^{j(\omega+\pi/4)})$$
Problem 8.14: Quadrature Multiplexing?✓

$y(t) = x_1\cos\omega_c t + x_2\sin\omega_c t$. Recover $x_1, x_2$.

Solution

Multiply by $\cos\omega_c t$ + LPF $\Rightarrow \frac{1}{2}x_1(t)$. Multiply by $\sin\omega_c t$ + LPF $\Rightarrow \frac{1}{2}x_2(t)$.
Works because $\cos$ and $\sin$ are orthogonal.

Problem 8.15: PCM Bit Rate?✓

4 kHz audio, $f_s = 8$ kHz, 8 bits/sample.

Solution
$$\text{Bit rate} = f_s \times \text{bits/sample} = 8000\times 8 = 64\;\text{kbps}$$

Chapter 9: Laplace Transform

Problem 9.2: Two-Sided Signal?✓

$x(t) = e^{-2t}u(t) - e^{3t}u(-t)$.

Solution
$$X(s) = \frac{1}{s+2} + \frac{1}{s-3} = \frac{2s-1}{(s+2)(s-3)}, \qquad \text{ROC: } -2 < \text{Re}\{s\} < 3$$

ROC includes $j\omega$ axis $\Rightarrow$ Fourier transform exists.

Problem 9.10: $s$-Domain Differentiation?✓

$X(s) = \frac{1}{s+3}$. Laplace transform of $t\,x(t)$.

Solution
$$t\,x(t) \;\longleftrightarrow\; -\frac{dX(s)}{ds} = -\frac{d}{ds}\left[\frac{1}{s+3}\right] = \frac{1}{(s+3)^2}$$
Problem 9.13: 2nd-Order Butterworth?✓

Find $H(s)$ for 2nd-order Butterworth with $\omega_c=1$.

Solution
$$|H(j\omega)|^2 = \frac{1}{1+\omega^4} \;\Longrightarrow\; \boxed{H(s) = \frac{1}{s^2+\sqrt{2}\,s+1}}$$

Poles at $s = \frac{-1\pm j}{\sqrt{2}}$.

Problem 9.14: Time Scaling?✓

$x(t) \longleftrightarrow X(s)$ with ROC $\mathcal{R}$. Find Laplace transform of $x(3t)$.

Solution
$$x(at) \;\longleftrightarrow\; \frac{1}{|a|}X\!\left(\frac{s}{a}\right), \quad\text{ROC: } \{s : s/a \in \mathcal{R}\}$$
Problem 9.15: Parallel Systems?✓

$H_1(s)=\frac{1}{s+1}$, $H_2(s)=\frac{1}{s+2}$ in parallel.

Solution
$$H(s) = \frac{1}{s+1}+\frac{1}{s+2} = \frac{2s+3}{(s+1)(s+2)}$$
$$h(t) = \bigl(e^{-t}+e^{-2t}\bigr)\,u(t)$$

Chapter 10: z-Transform

Problem 10.8: Geometric Evaluation?✓

$H(z) = \frac{z-1}{z-0.9}$. Evaluate $|H(e^{j\omega})|$ at $\omega=0$ and $\omega=\pi$.

Solution
$$|H(e^{j0})| = \frac{|1-1|}{|1-0.9|} = \frac{0}{0.1} = 0$$
$$|H(e^{j\pi})| = \frac{|-1-1|}{|-1-0.9|} = \frac{2}{1.9} \approx 1.053$$

Zero at $\omega=0$ $\Rightarrow$ highpass filter.

Problem 10.9: Pole-Zero Cancellation?✓

$H(z) = \frac{z-0.5}{(z-0.5)(z-0.8)}$.

Solution
$$H(z) = \frac{1}{z-0.8} = \frac{z^{-1}}{1-0.8z^{-1}} \;\Longrightarrow\; h[n] = (0.8)^{n-1}u[n-1]$$
Problem 10.10: $z$-Domain Differentiation?✓

$x[n] = n(0.5)^n u[n]$.

Solution
$$n\,x[n] \;\longleftrightarrow\; -z\frac{dX(z)}{dz}$$
$$\mathcal{Z}\{n(0.5)^n u[n]\} = \frac{0.5z^{-1}}{(1-0.5z^{-1})^2}, \qquad \text{ROC: } |z|>0.5$$
Problem 10.12: Inverse System?✓

$H(z) = \frac{1-0.5z^{-1}}{1-0.8z^{-1}}$. Find inverse system.

Solution
$$H_{\text{inv}}(z) = \frac{1}{H(z)} = \frac{1-0.8z^{-1}}{1-0.5z^{-1}}$$

Pole at $z=0.5$ (inside unit circle) $\Rightarrow$ causal inverse is stable.

Problem 10.13: Cascade Decomposition?✓

$H(z) = \frac{1+z^{-1}}{(1-0.5z^{-1})(1+0.3z^{-1})}$.

Solution
$$H(z) = \underbrace{\frac{1+z^{-1}}{1-0.5z^{-1}}}_{H_1(z)}\cdot\underbrace{\frac{1}{1+0.3z^{-1}}}_{H_2(z)}$$