Random Questions Signals & Systems

Fourier Transform

Fourier Transform 1

To compute the Fourier Transform of the given signal:

The signal is:

$x(t)=\delta (t+2)+\delta (t-2)+5\delta (t+1)-5\delta (t-1)x(t)$ $=\delta (t+2)+\delta (t-2)+5\delta (t+1)-5\delta (t-1)$

Fourier Transform of Dirac Delta Functions

The Fourier Transform of $\delta (t-a)$ is given by: $\mathcal{F}\{\delta (t-a)\}=e^{-j\omega a}$

where $a$ is the shift in time.

Fourier Transform of $x(t)$

Using the linearity property of the Fourier Transform, we compute the Fourier Transform of each term:

1. $\mathcal{F}\delta (t+2)=e^{j2\omega }$
2. $\mathcal{F}\delta (t-2)=e^{-j2\omega }$
3. $\mathcal{F}5\delta (t+1)=5e^{j\omega }$
4. $\mathcal{F}-5\delta (t-1)=-5e^{-j\omega }$

Adding these together, the Fourier Transform of $x(t)$ is:

$X(\omega )=ej2\omega +e-j2\omega +5ej\omega -5e-j\omega X(\omega )=e^{j2\omega }+e^{-j2\omega }+5e^{j\omega }-5e^{-j\omega }$

Simplifying Using Euler’s Formula

Using Euler’s formula:

$=ej\theta +e-j\theta$ $=2cos(\theta )e^{j\theta }+e^{-j\theta }$ $=2\cos (\theta )ej\theta -e-j\theta$ $=2jsin(\theta )e^{j\theta }-e^{-j\theta }$ $=2j\sin (\theta )$

We simplify:

1. $e^{j2\omega }+e^{-j2\omega }=2\cos (2\omega )$
2. $5e^{j\omega }-5e^{-j\omega }=10j\sin (\omega )$

Thus, the Fourier Transform is:

$X(\omega )=2cos(2\omega )+10jsin(\omega )X(\omega )=2\cos (2\omega )+10j\sin (\omega )$

This is the final expression for the Fourier Transform of $x(t)$.

Fourier Transform 2

Here is the solution written in Obsidian-friendly format:

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Solution

The given signal is:

x(t)={e−at,t>0−eat,t<0x(t) = \begin{cases} e^{-at}, & t > 0 \ -e^{at}, & t < 0 \end{cases}

where $a>0$.

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Fourier Transform Definition

The Fourier Transform is given by:

X(jω)=∫−∞∞x(t)e−jωtdtX(j\omega) = \int_{-\infty}^\infty x(t) e^{-j\omega t} dt

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Splitting the Signal

Split the Fourier Transform into two parts based on the given signal:

X(jω)=∫−∞0−eate−jωtdt+∫0∞e−ate−jωtdtX(j\omega) = \int_{-\infty}^0 -e^{at} e^{-j\omega t} dt + \int_{0}^\infty e^{-at} e^{-j\omega t} dt

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Step 1: Solve for $t>0$

For $t>0$, $x(t)=e^{-at}$:

∫0∞e−ate−jωtdt=∫0∞e−(a+jω)tdt\int_{0}^\infty e^{-at} e^{-j\omega t} dt = \int_{0}^\infty e^{-(a + j\omega)t} dt

Using the formula ${\int }_0^{\infty }{e}^{-kt}dt=\frac{1}{k}$ for $\text{Re}(k)>0$:

∫0∞e−(a+jω)tdt=1a+jω\int_{0}^\infty e^{-(a + j\omega)t} dt = \frac{1}{a + j\omega}

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Step 2: Solve for $t<0$

For $t<0$, $x(t)=-e^{at}$:

∫−∞0−eate−jωtdt=−∫−∞0e(a−jω)tdt\int_{-\infty}^0 -e^{at} e^{-j\omega t} dt = -\int_{-\infty}^0 e^{(a - j\omega)t} dt

Substitute $t\to -t$ to handle the bounds:

−∫−∞0e(a−jω)tdt=∫0∞e−(a−jω)tdt-\int_{-\infty}^0 e^{(a - j\omega)t} dt = \int_{0}^\infty e^{-(a - j\omega)t} dt

Using the same formula:

∫0∞e−(a−jω)tdt=1a−jω\int_{0}^\infty e^{-(a - j\omega)t} dt = \frac{1}{a - j\omega}

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Step 3: Combine the Results

Combine the results for $t>0$ and $t<0$:

X(jω)=1a+jω+1a−jωX(j\omega) = \frac{1}{a + j\omega} + \frac{1}{a - j\omega}

Simplify:

X(jω)=a−jω+a+jω(a+jω)(a−jω)X(j\omega) = \frac{a - j\omega + a + j\omega}{(a + j\omega)(a - j\omega)} X(jω)=2aa2+ω2X(j\omega) = \frac{2a}{a^2 + \omega^2}

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Magnitude Spectrum

The magnitude of $X(j\omega )$ is:

∣X(jω)∣=2aa2+ω2|X(j\omega)| = \frac{2a}{a^2 + \omega^2}

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Phase Spectrum

The phase of $X(j\omega )$ is:

∠X(jω)=0,since the signal is real and even.\angle X(j\omega) = 0, \quad \text{since the signal is real and even.}

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Final Answer

1. Fourier Transform:

X(jω)=2aa2+ω2X(j\omega) = \frac{2a}{a^2 + \omega^2}

2. Magnitude Spectrum:

∣X(jω)∣=2aa2+ω2|X(j\omega)| = \frac{2a}{a^2 + \omega^2}

3. Phase Spectrum:

∠X(jω)=0\angle X(j\omega) = 0

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The magnitude spectrum is a Lorentzian curve centered at $\omega =0$ with a peak value of $\frac{2}{a}$. The phase spectrum is constant at $0$. Let me know if you’d like additional visualizations! h

Z Transform

Question 1

Question:

Compute the Z-transform and sketch the ROC for ( x[n] = {5, 4, -3, -1, 0, 4, -3, 2} ), where ( n = -3, -2, -1, 0, 1, 2, 3, 4 ).

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Solution:

1. Z-transform:

   $$X(z)=5z^3+4z^2-3z-1+\frac{4}{z^2}-\frac{3}{z^3}+\frac{2}{z^4}$$

2. Region of Convergence (ROC):

   $$\text{ROC: All values of }z,\text{ except }z=0.$$

Question 2

Question:

Calculate the Z-transform of the following discrete-time signals and also sketch the ROC:

1. $x[n]={\left(\frac{1}{4}\right)}^{n}u[n]+{\left(\frac{1}{5}\right)}^{n}u[-n-1]$
2. $x[n]=\sin ({\omega }_{0}n)u[n]$

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Solution:

(i) $x[n]={\left(\frac{1}{4}\right)}^{n}u[n]+{\left(\frac{1}{5}\right)}^{n}u[-n-1]$

1. For ${\left(\frac{1}{4}\right)}^{n}u[n]$:

   The Z-transform is:

   $$X_1(z)=\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n{z}^{-n}$$

   This is a geometric series with the first term $1$ and the common ratio $\frac{1}{4z}$:

   X_1(z) = \frac{1}{1 - \frac{1}{4}z^{-1}} = \frac{z}{z - \frac{1}{4}}. $$ **ROC**: $|z| > \frac{1}{4}$.

2. For ${\left(\frac{1}{5}\right)}^{n}u[-n-1]$:

   Rewrite the summation as $n=-k$, where $k\ge 1$, giving:

   $$X_2(z)=\sum _{k=1}^{\infty }{\left(\frac{1}{5}\right)}^k{z}^{-k}.$$

   This is a geometric series with the first term $\frac{1}{5z}$ and the common ratio $\frac{1}{5z}$:

   $$X_2(z)=\frac{\frac{1}{5z}}{1-\frac{1}{5z}}=\frac{\frac{1}{5}}{z-\frac{1}{5}}.$$

   ROC: $|z|<\frac{1}{5}$.

3. Total Z-transform:

   $$X(z)=X_1(z)+X_2(z)=\frac{z}{z-\frac{1}{4}}+\frac{\frac{1}{5}}{z-\frac{1}{5}}.$$

   Total ROC: No overlap in ROCs, so the Z-transform is valid only in disjoint regions:

   • $|z|>\frac{1}{4}$ for $X_1(z)$.
   • $|z|<\frac{1}{5}$ for $X_2(z)$.

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(ii) $x[n]=\sin ({\omega }_{0}n)u[n]$

1. Rewrite $\sin ({\omega }_{0}n)$ using Euler’s formula:

   $$\sin ({\omega }_{0}n)=\frac{e^{j{\omega }_{0}n}-e^{-j{\omega }_{0}n}}{2j}.$$

   The signal becomes:

   $$x[n]=\frac{1}{2j}\left(e^{j{\omega }_{0}n}-e^{-j{\omega }_{0}n}\right)u[n].$$

2. Compute the Z-transform for each term separately:

   • For $e^{j{\omega }_{0}n}u[n]$:

     $$X_1(z)=\sum _{n=0}^{\infty }{e}^{j{\omega }_{0}n}{z}^{-n}=\frac{1}{1-e^{j{\omega }_0}{z}^{-1}}.$$

   • For $e^{-j{\omega }_{0}n}u[n]$:

     $$X_2(z)=\sum _{n=0}^{\infty }{e}^{-j{\omega }_{0}n}{z}^{-n}=\frac{1}{1-e^{-j{\omega }_0}{z}^{-1}}.$$

3. Combine terms:

   $$X(z)=\frac{1}{2j}\left(\frac{1}{1-e^{j{\omega }_0}{z}^{-1}}-\frac{1}{1-e^{-j{\omega }_0}{z}^{-1}}\right).$$

4. Simplify:

   $$X(z)=\frac{z^{-1}\sin ({\omega }_0)}{1-2\cos ({\omega }_0)z^{-1}+z^{-2}}.$$

   ROC: $|z|>1$ (signal is causal).

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Final Answers:

1. For $x[n]={\left(\frac{1}{4}\right)}^{n}u[n]+{\left(\frac{1}{5}\right)}^{n}u[-n-1]$:

   • $X(z)=\frac{z}{z-\frac{1}{4}}+\frac{\frac{1}{5}}{z-\frac{1}{5}}$
   • ROC: Disjoint regions, $|z|>\frac{1}{4}$ and $|z|<\frac{1}{5}$.

2. For $x[n]=\sin ({\omega }_{0}n)u[n]$:

   • $X(z)=\frac{z^{-1}\sin ({\omega }_0)}{1-2\cos ({\omega }_0)z^{-1}+z^{-2}}$
   • ROC: $|z|>1$.

Time Period

Question 1

Question:

Compute the fundamental time period and frequency of the following signals:

1. $x(t)=5\sin (24\pi t)+7\sin (36\pi t)$
2. $x(t)=5\cos (\pi t)\sin (3\pi t)$

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Solution:

(i) $x(t)=5\sin (24\pi t)+7\sin (36\pi t)$

1. The angular frequency of $\sin (24\pi t)$ is ${\omega }_1=24\pi$, and the period is:

   $$T_1=\frac{2\pi }{{\omega }_1}=\frac{2\pi }{24\pi }=\frac{1}{12}.$$

2. The angular frequency of $\sin (36\pi t)$ is ${\omega }_2=36\pi$, and the period is:

   $$T_2=\frac{2\pi }{{\omega }_2}=\frac{2\pi }{36\pi }=\frac{1}{18}.$$

3. The fundamental time period $T$ is the least common multiple (LCM) of $T_1$ and $T_2$:

   $$T=\text{LCM}\left(\frac{1}{12},\frac{1}{18}\right)=\frac{1}{6}.$$

4. The fundamental frequency $f$ is:

   $$f=\frac{1}{T}=6\text{Hz}.$$

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(ii) $x(t)=5\cos (\pi t)\sin (3\pi t)$

1. Using the trigonometric identity:

   $$\cos (A)\sin (B)=\frac{1}{2}\left[\sin (A+B)-\sin (A-B)\right],$$

   the signal becomes:

   $$x(t)=\frac{5}{2}\left[\sin (4\pi t)-\sin (2\pi t)\right].$$

2. The angular frequencies are:

   • For $\sin (4\pi t)$, ${\omega }_1=4\pi$, and $T_1=\frac{1}{2}.$
   • For $\sin (2\pi t)$, ${\omega }_2=2\pi$, and $T_2=1.$

3. The fundamental time period $T$ is the LCM of $T_1$ and $T_2$:

   $$T=\text{LCM}\left(\frac{1}{2},1\right)=1.$$

4. The fundamental frequency $f$ is:

   $$f=\frac{1}{T}=1\text{Hz}.$$

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Final Answers:

1. For $x(t)=5\sin (24\pi t)+7\sin (36\pi t)$:

   • Fundamental Time Period: $T=\frac{1}{6}\text{s}$
   • Frequency: $f=6\text{Hz}$

2. For $x(t)=5\cos (\pi t)\sin (3\pi t)$:

   • Fundamental Time Period: $T=1\text{s}$
   • Frequency: $f=1\text{Hz}$

Question 2

Convolution

Question:

Compute the convolution of the given two signals:

• $x(t)=u(t)-u(t-4)$
• $h(t)=u(t)-u(t-1)$

using the graphical method.

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Solution:

1. Signals Description:

1. $x(t)=u(t)-u(t-4)$: This represents a rectangular pulse starting at $t=0$ and ending at $t=4$.
2. $h(t)=u(t)-u(t-1)$: This represents a rectangular pulse starting at $t=0$ and ending at $t=1$.

2. Convolution Formula:

The convolution of two signals $x(t)$ and $h(t)$ is given by:

$$y(t)=(x\ast h)(t)={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau$$

3. Graphical Method Steps:

1. Flip $h(t)$ to obtain $h(-\tau )$:

   • $h(-\tau )$ is a pulse of width 1 (from $\tau =-1$ to $\tau =0$).

2. Shift $h(-\tau )$ to $h(t-\tau )$:

   • As $t$ varies, $h(t-\tau )$ shifts across $t$.

3. Multiply $x(\tau )$ and $h(t-\tau )$:

   • For each value of $t$, calculate the overlap between $x(\tau )$ (a rectangular pulse from $0$ to $4$) and $h(t-\tau )$ (a rectangular pulse shifted to $t$).

4. Integrate the product:

   • The integral computes the area of overlap between the two pulses.

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4. Piecewise Convolution Results:

We evaluate the convolution $y(t)$ in different time intervals based on the overlap of $x(\tau )$ and $h(t-\tau )$:

1. For $t<0$:

   • No overlap between $x(\tau )$ and $h(t-\tau )$.
   • $y(t)=0$.

2. For $0\le t<1$:

   • $h(t-\tau )$ starts to overlap with $x(\tau )$ from $\tau =0$ to $\tau =t$.
   • Overlap area = width of $h(t-\tau )$ = $t$.
   • $y(t)=t$.

3. For $1\le t<4$:

   • Full overlap between $h(t-\tau )$ and $x(\tau )$ from $\tau =0$ to $\tau =1$.
   • Overlap area = width of $h(t-\tau )$ = $1$.
   • $y(t)=1$.

4. For $4\le t<5$:

   • $h(t-\tau )$ overlaps partially with $x(\tau )$ from $\tau =t-4$ to $\tau =1$.
   • Overlap area = width = $5-t$.
   • $y(t)=5-t$.

5. For $t\ge 5$:

   • No overlap between $x(\tau )$ and $h(t-\tau )$.
   • $y(t)=0$.

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5. Final Expression for $y(t)$:

The result of the convolution is:

$$y(t)=\{\begin{array}{ll}0,& t<0\\ t,& 0\le t<1\\ 1,& 1\le t<4\\ 5-t,& 4\le t<5\\ 0,& t\ge 5\end{array}$$

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Graphical Representation:

1. The convolution $y(t)$ is piecewise linear and can be plotted:
   • Rising linearly from $0$ to $1$ in the interval $[0,1]$.
   • Constant at $1$ in the interval $[1,4]$.
   • Falling linearly from $1$ to $0$ in the interval $[4,5]$.

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Let me know if you would like me to plot the graph for this! You can copy-paste this explanation into Obsidian.

Energy Power Signal

Question 1

Question:

Classify the following signals as energy signal, power signal, or neither. Also, determine the energy and power.

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

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Solution:

Definitions:

1. Energy Signal: A signal is an energy signal if its total energy is finite and power is zero:

   $$E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt<\infty ,P=0.$$

2. Power Signal: A signal is a power signal if its average power is finite and energy is infinite:

   $$P=\underset{T\to \infty }{\lim }\frac{1}{2T}{\int }_{-T}^T|x(t){|}^{2}dt<\infty ,E=\infty .$$

3. Neither: If neither condition is satisfied, the signal is classified as “neither.”

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

1. Signal Analysis:

   • The term $u(-t)$ ensures that $x(t)$ is nonzero only for $t\le 0$.
   • For $t\le 0$, $x(t)=e^{3t}$.

2. Energy Calculation:

   $$E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt={\int }_{-\infty }^0{\left(e^{3t}\right)}^{2}dt={\int }_{-\infty }^0{e}^{6t}dt.$$

   Solve the integral:

   $$E={\left[\frac{e^{6t}}{6}\right]}_{-\infty }^0=\frac{1}{6}\left(e^0-\underset{t\to -\infty }{\lim }{e}^{6t}\right)=\frac{1}{6}.$$
   • The energy is finite: $E=\frac{1}{6}$.

3. Power Calculation: The power is zero since the signal exists only for $t\le 0$ (non-periodic and finite energy).

4. Classification:

   • Energy Signal.

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(ii) $x(t)=e^{-2|t|}$

1. Signal Analysis:

   • $x(t)$ is symmetric around $t=0$.
   • For $t\ge 0$, $x(t)=e^{-2t}$.
   • For $t<0$, $x(t)=e^{2t}$.

2. Energy Calculation:

   $$E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt={\int }_{-\infty }^0{\left(e^{2t}\right)}^{2}dt+{\int }_0^{\infty }{\left(e^{-2t}\right)}^{2}dt.$$

   Solve for each integral:

   • For $t\ge 0$: ${\int }_0^{\infty }{e}^{-4t}dt=\frac{1}{4}$.
   • For $t<0$: ${\int }_{-\infty }^0{e}^{4t}dt=\frac{1}{4}$.

   Total energy:

   $$E=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$$
   • The energy is finite: $E=\frac{1}{2}$.

3. Power Calculation: The power is zero since the signal is non-periodic and finite energy.

4. Classification:

   • Energy Signal.

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Final Answers:

1. For $x(t)=e^{3t}u(-t)$:

   • Classification: Energy Signal.
   • Energy: $E=\frac{1}{6}$.
   • Power: $P=0$.

2. For $x(t)=e^{-2|t|}$:

   • Classification: Energy Signal.
   • Energy: $E=\frac{1}{2}$.
   • Power: $P=0$.

References

Information

• date: 2024.11.25
• time: 20:36