4.1 HW 1

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4.1.1 Problem 1.8, Chapter 1

Express the real part of each of the following signals in the form $A{e}^{-at}\cos (\omega t+\varphi )$, where $A,a,\omega ,\varphi$ are real numbers with $A>0$ and $-\pi <\varphi \le \pi$: (a) $x_1\left(t\right)=-2$, (b) $x_2\left(t\right)=\sqrt{2}{e}^{j\frac{\pi }{4}}\cos (3t+2\pi )$, (c) $x_3\left(t\right)=e^{-t}\sin (3t+\pi )$, (d) $x_4\left(t\right)=j{e}^{(-2+100j)t}$

Solution

4.1.1.1 part a

$$x_1\left(t\right)=-2$$ Comparing the above to $A{e}^{-at}\cos (\omega t+\varphi )$ shows that $$A=2,a=0,\varphi =0,\omega =0,\varphi =-\pi$$

4.1.1.2 part b

$$x_2\left(t\right)=\sqrt{2}{e}^{j\frac{\pi }{4}}\cos (3t+2\pi )$$ Since $\cos (3t+2\pi )=\cos \left(3t\right)$, the above becomes$$\begin{array}{rl}{x}_2\left(t\right)& =\sqrt{2}{e}^{j\frac{\pi }{4}}\cos \left(3t\right)\\ & =\sqrt{2}(\cos \frac{\pi }{4}+j\sin \frac{\pi }{4})\cos \left(3t\right)\\ & =\sqrt{2}(\frac{1}{2}\sqrt{2}+j\frac{1}{2}\sqrt{2})\cos \left(3t\right)\\ & =(1+j)\cos \left(3t\right)\end{array}$$

Hence the real part of $x_2\left(t\right)$ is $$\mathrm{Re}\left(x_2\left(t\right)\right)=\cos \left(3t\right)$$ Comparing the above to $A{e}^{-at}\cos (\omega t+\varphi )$ shows that $$A=1,a=0,\omega =3,\varphi =0$$

4.1.1.3 part c

$$x_3\left(t\right)=e^{-t}\sin (3t+\pi )$$ Since $\sin (3t+\pi )=\cos (3t+\pi -\frac{\pi }{2})=\cos (3t+\frac{\pi }{2})$, then the above becomes$$x_3\left(t\right)=e^{-t}\cos (3t+\frac{\pi }{2})$$ Comparing the above to $A{e}^{-at}\cos (\omega t+\varphi )$ shows that $$A=1,a=1,\omega =3,\varphi =\frac{\pi }{2}$$

4.1.1.4 part d

$$\begin{array}{rl}{x}_4\left(t\right)& =j{e}^{(-2+100j)t}\\ & =j{e}^{-2t}{e}^{j100t}\end{array}$$

But $j=e^{j\frac{\pi }{2}}$, hence the above becomes$$\begin{array}{rl}{x}_4\left(t\right)& =e^{j\frac{\pi }{2}}{e}^{-2t}{e}^{j100t}\\ & =e^{-2t}{e}^{j(100t+\frac{\pi }{2})}\\ & =e^{-2t}(\cos (100t+\frac{\pi }{2})+j\sin (100t+\frac{\pi }{2}))\end{array}$$

Therefore $$\mathrm{Re}\left(x_4\left(t\right)\right)=e^{-2t}\cos (100t+\frac{\pi }{2})$$ Comparing the above to $A{e}^{-at}\cos (\omega t+\varphi )$ shows that $$A=1,a=2,\omega =100,\varphi =\frac{\pi }{2}$$

4.1.2 Problem 1.13, Chapter 1

Consider the continuous-time signal $x\left(t\right)=\delta (t+2)-\delta (t-2)$. Calculate the value of $E_{\mathrm{\infty }}$ for the signal $y\left(t\right)={\int }_{-\mathrm{\infty }}^{t}x\left(\tau \right)d\tau$

Solution

$y\left(t\right)$ is first found$$\begin{array}{rl}y\left(t\right)& ={\int }_{-\mathrm{\infty }}^t\delta (t+2)-\delta (t-2)d\tau \\ & ={\int }_{-\mathrm{\infty }}^t\delta (t+2)d\tau -{\int }_{-\mathrm{\infty }}^t\delta (t-2)d\tau \end{array}$$

$\delta (t+2)$ is an impulse at $t=-2$ and $\delta (t-2)$ is an impulse at $t=2$. Hence is $t<-2$ then the above is zero. If $-2<t<2$ then only the first integral contributes giving $1$ and if $t>2$ then both integral contribute $1$ each, and hence cancel each others giving $y=0$. Therefore$$y\left(t\right)=\{\begin{array}{cc}0& t<-2\\ 1& -2<t<2\\ 0& t>2\end{array}$$ Now that $y\left(t\right)$ is found, its $E_{\mathrm{\infty }}$ can be calculated using the definition $$\begin{array}{rl}{E}_{\mathrm{\infty }}& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|y\left(t\right)\right|}^{2}dt\\ & ={\int }_{-2}^{2}1dt\\ & ={\left[t\right]}_{-2}^2\\ & =2+2\end{array}$$

Hence$$E_{\mathrm{\infty }}=4$$

4.1.3 Problem 1.17, Chapter 1

Consider a continuous-time system with input $x(t)$ and output $y(t)$ related by $y\left(t\right)=x(\sin \left(t\right))$. (a) Is this system causal? (b) Is this system linear?

Solution

4.1.3.1 Part a

A system is causal if its output at time $t$ depends only on current $t$ and on past $t$ and not on future $t$. Picking $t=-\pi$, then $y(-\pi )=x(\sin (-\pi ))=x\left(0\right)$. This shows that $y(-\pi )=x\left(0\right)$. Hence the output depends on input at future time (since $0>-\pi$). Therefore this system is not causal.

4.1.3.2 Part b

Let input be $x\left(t\right)=a_1{x}_1\left(t\right)+a_2{x}_2\left(t\right)$. If the output when the input is $x\left(t\right)$ is given by $y\left(t\right)=a_1{y}_1\left(t\right)+a_2{y}_2\left(t\right)$ where $y_1\left(t\right)=x_1\left(t\right)$ and $y_2\left(t\right)=x_2\left(t\right)$ then the system is linear.  From the definition$$\begin{array}{rl}y\left(t\right)& =x(\sin \left(t\right))\\ & =a_1{x}_1(\sin \left(t\right))+a_2{x}_2(\sin \left(t\right))\end{array}$$

Now, $y_1\left(t\right)=x_1(\sin t)$ and $y_2\left(t\right)=x_2(\sin t)$. Hence the above becomes$$\begin{array}{rl}y\left(t\right)& =x(\sin \left(t\right))\\ & =a_1{y}_1\left(t\right)+a_2{y}_2\left(t\right)\end{array}$$

Therefore the system is linear.

4.1.4 Problem 1.21, Chapter 1

A continuous-time signal $x(t)$ is shown in Figure P1.21. Sketch and label carefully each of the following signals: (a) $x(t-1)$. (b) $x(2-t)$ (c) $x(2t+1)$ (d) $x(4-\frac{t}{2})$

Figure 4.1:The function $x(t)$

Solution

Looking at the plot, it can be constructed from unit step $u\left(t\right)$ and ramp function $r\left(t\right)$ as follows$$x\left(t\right)=-u(t+2)+r(t+2)-r(t+1)+u(t+1)+u\left(t\right)-u(t-1)-r(t-1)+r(t-2)$$ Here is an implementation

Figure 4.2:Construction the signal $x(t)$ from unit step and ramp functions

Figure 4.3:Code for the above

4.1.4.1 Part a

$x(t-1)$ is $x\left(t\right)$ shifted to right by one unit time. Hence it becomes$$x(t-1)=-u(t+1)+r(t+1)-r\left(t\right)+u\left(t\right)+u(t-1)-u(t-2)-r(t-2)+r(t-3)$$

Figure 4.4:Part (a) plot

Figure 4.5:Code for the above

4.1.4.2 Part b

$$x(2-t)=x(-(t-2))$$ Hence the signal $x\left(t\right)$ is first flipped right to left (also called reflection about the $t=0$ axis) and the resulting function is then shifted to the right by 2 units. It becomes

$$\begin{array}{rl}x(2-t)& =-u((2-t)+2)+r((2-t)+2)-r((2-t)+1)+u((2-t)+1)+u(2-t)-u((2-t)-1)-r((2-t)-1)+r((2-t)-2)\\ & =-u(4-t)+r(4-t)-r(3-t)+u(3-t)+u(2-t)-u(1-t)-r(1-t)+r(-t)\end{array}$$

The flipped signal is

Figure 4.6:Part (b) signal after reflection

Now the above is shifted to the right by 2 units giving

Figure 4.7:Part (b) final plot

It also possible to first do the shifting, followed by the reflection. Same output will result.

4.1.4.3 Part c

$x(2t+1)=x\left(2(t+\frac{1}{2})\right)$. The signal is first shifted to the left by $\frac{1}{2}$ due to the $+\frac{1}{2}$ term, and then the resulting signal is squashed (contraction) by factor of $2$. Since the original signal is from $-1$ to $3$, then after first shifting it to the left by $0.5$ it becomes from $-1.5$ to $2.5$. Hence the original ramp that went from $-2$ to $-1$ now goes from $-1.5$ to $-1$ and the line that originally went from $-1$ to $0$ now goes from $-1$ to $-\frac{1}{2}$ (half the length) and so on. This is the result

Figure 4.8:Part (c) plot

Figure 4.9:Code for the above

4.1.4.4 Part d

$x(4-\frac{t}{2})=x(-(\frac{t}{2}-4))=x(-\frac{1}{2}(t-8))$. Hence, the signal is first shifted to the right by $8$ due to the $-8$ term, and then the resulting signal is flipped across the $t=0$ axis, and then the resulting signal is stretched (expanded) by factor of $2$ due to the multiplication by $\frac{1}{2}$ term. This is the result showing each step

Figure 4.10:Part (d) plot

4.1.5 Problem 1.22, Chapter 1

A discrete-time signal is shown in Figure P1.22. Sketch and label carefully each of the following signals (a) $x[n-4]$ (b) $x[3-n]$ (c) $x\left[3n\right]$ (d) $x[3n+1]$

Figure 4.11:The function $x[n]$

Solution

4.1.5.1 Part a

$x[n-4]$ is $x\left[n\right]$ shifted to the right by $4$ positions. Hence it becomes

Figure 4.12:Part (a) plot

Figure 4.13:Code used for the above

4.1.5.2 Part b

$x[3-n]=x[-(n-3)]$. Hence $x[n]$ is first reflected to obtain $x[-n]$ and then the result is shifted to right by $3$. This is the result showing each step

Figure 4.14:Part (b) plot

4.1.5.3 Part c

$x\left[3n\right]$. Sample at $n=0$ remains the same. Sample at $n=-1$ gets the value of the sample that was at $-3$ which is $-\frac{1}{2}$. Sample at $n=-2$ gets the value of the sample that was at $n=-6$ which is zero. Hence for all $n$ less than $-1$ new values are all zero. Same for the right side. The sample at $n=1$ gets the value of the sample that was at $3$ which is $\frac{1}{2}$ and sample at $n=2$ gets the value of the sample that was at $6$ which is $0$ and all $n>1$ are therefore zero. Notice that this operation causes samples to be lost from the original signal. This is the final result

Figure 4.15:Part (c) plot

4.1.5.4 Part d

$x[3n+1]$. Sample at $n=0$ gets the value that was at $n=1$ which is $1$. Sample at $n=-1$ gets the value of the sample that was at $-3+1=-2$ which is $\frac{1}{2}$. Sample at $n=-2$ gets the value of the sample that was at $n=-6+1=-5$ which is zero. Hence for all $n<-1$ all values are zero. Same for the right side. Sample at $n=1$ gets the value of the sample that was at $3+1=4$ which is $0$ and therefore for all $n>1$ samples are zero. Notice that this operation causes samples to be lost from the original signal. The result is

Figure 4.16:Part (d) plot

4.1.6 Problem 1.26, Chapter 1

Determine whether or not each of the following discrete-time signals is periodic. If the signal is periodic, determine its fundamental period (a) $\sin (\frac{6}{7}\pi n+1)$ (b) $x\left[n\right]=\cos (\frac{n}{8}-\pi )$ (c) $x\left[n\right]=\cos \left(\frac{\pi }{8}{n}^2\right)$ (d) $x\left[n\right]=\cos \left(\frac{\pi }{2}n\right)\cos \left(\frac{\pi }{4}n\right)$

Solution

The signal $x\left[n\right]$ is periodic, if integer $N$ can be found that $x\left[n\right]=x[n+N]$ for all $n$. Fundamental period is the smallest such integer $N$.

4.1.6.1 Part a

In this part, $x\left[n\right]=\sin (\frac{6}{7}\pi n+1)$. Hence the signal is periodic if $$\begin{array}{rl}x\left[n\right]& =x[n+N]\\ \sin (\frac{6}{7}\pi n+1)& =\sin (\frac{6}{7}\pi (n+N)+1)\\ & =\sin ((\frac{6}{7}\pi n+1)+\frac{6}{7}\pi N)\end{array}$$

The above will be true if $$\frac{6}{7}\pi N=2\pi m$$ For some integer $m$ and $N$. This is because $\sin$ has $2\pi$ period. This implies that$$\frac{3}{7}=\frac{m}{N}$$ Therefore $N=7,m=3$. Since it was possible to find $n,N$ integers, then it is periodic. Since $m,N$ are relatively prime then $N=7$ is the fundamental period.

4.1.6.2 Part b

In this part, $x\left[n\right]=\cos (\frac{n}{8}-\pi )$. Hence the signal is periodic if $$\begin{array}{rl}x\left[n\right]& =x[n+N]\\ \cos (\frac{n}{8}-\pi )& =\cos (\frac{n+N}{8}-\pi )\\ & =\cos ((\frac{n}{8}-\pi )+\frac{N}{8})\end{array}$$

The above will be true if $$\begin{array}{rl}\frac{N}{8}& =2\pi m\\ \frac{1}{16\pi }& =\frac{m}{N}\end{array}$$

For some integer $N,m$.  It is not possible to find integers $m,N$ to satisfy the above since $\pi$ is an irrational number. Hence not periodic.

4.1.6.3 Part c

$x\left[n\right]=\cos \left(\frac{\pi }{8}{n}^2\right)$. Hence the signal is periodic if $$\begin{array}{rl}x\left[n\right]& =x[n+N]\\ \cos \left(\frac{\pi }{8}{n}^2\right)& =\cos \left(\frac{\pi }{8}{(n+N)}^2\right)\\ & =\cos \left(\frac{\pi }{8}(n^2+N^2+2nN)\right)\\ & =\cos (\frac{\pi }{8}{n}^2+\frac{\pi }{8}(N^2+2nN))\end{array}$$

The above will be true if $$\frac{\pi }{8}(N^2+2nN)=2\pi m$$ Need to find smallest integer $N$ to satisfy this for all $n$. Choosing $N=8$ the above becomes$$\begin{array}{rl}\frac{\pi }{8}(64+16n)& =m\left(2\pi \right)\\ 8\pi +2n\pi & =m\left(2\pi \right)\end{array}$$

Hence for all $n$, $N=8$ satisfies the equation (since $m$ is arbitrary integer). Therefore it is periodic  and fundamental period $N=8$.

4.1.6.4 Part d

Using $\cos A\cos B=\frac{1}{2}(\cos (A+B)+\cos (A-B))$ then$$\begin{array}{rl}\cos \left(\frac{\pi }{2}n\right)\cos \left(\frac{\pi }{4}n\right)& =\frac{1}{2}(\cos (\frac{\pi }{2}n+\frac{\pi }{4}n)+\cos (\frac{\pi }{2}n-\frac{\pi }{4}n))\\ & =\frac{1}{2}(\cos \left(\frac{3\pi }{4}n\right)+\cos \left(\frac{\pi }{4}n\right))\end{array}$$

Considering each signal separately. $x\left[n\right]=\cos \left(\frac{3\pi }{4}n\right)$. This is periodic if $$\begin{array}{rl}x\left[n\right]& =x[n+N]\\ \cos \left(\frac{3\pi }{4}n\right)& =\cos \left(\frac{3\pi }{4}(n+N)\right)\\ & =\cos (\left(\frac{3\pi }{4}n\right)+\frac{3\pi }{4}N)\end{array}$$

The above will be true if $$\begin{array}{rl}\frac{3\pi }{4}N& =2\pi m\\ \frac{3}{8}& =\frac{m}{N}\end{array}$$

It was possible to find integers $N,m$ to satisfy this, where period $N=8$. Considering the second signal $x\left[n\right]=\cos \left(\frac{\pi }{4}n\right)$. This is periodic if $$\begin{array}{rl}x\left[n\right]& =x[n+N]\\ \cos \left(\frac{\pi }{4}n\right)& =\cos \left(\frac{\pi }{4}(n+N)\right)\\ & =\cos (\left(\frac{\pi }{4}n\right)+\frac{\pi }{4}N)\end{array}$$

The above will be true if $$\begin{array}{rl}\frac{\pi }{4}N& =2\pi m\\ \frac{1}{8}& =\frac{m}{N}\end{array}$$

It was possible to find integers $N,m$ to satisfy this, where period $N=8$. Therefore both signals periodic with same period, the sum is therefore periodic and the fundamental period is $N=8$.

4.1.7 key solution

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