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 \(Ae^{-at}\cos \left ( \omega t+\phi \right ) \), where \(A,a,\omega ,\phi \) are real numbers with \(A>0\) and \(-\pi <\phi \leq \pi \): (a) \(x_{1}\left ( t\right ) =-2\), (b) \(x_{2}\left ( t\right ) =\sqrt{2}e^{j\frac{\pi }{4}}\cos \left ( 3t+2\pi \right ) \), (c) \(x_{3}\left ( t\right ) =e^{-t}\sin \left ( 3t+\pi \right ) \), (d) \(x_{4}\left ( t\right ) =je^{\left ( -2+100j\right ) t}\)

Solution

4.1.1.1 part a

\[ x_{1}\left ( t\right ) =-2 \] Comparing the above to \(Ae^{-at}\cos \left ( \omega t+\phi \right ) \) shows that \[ A=2,a=0,\phi =0,\omega =0,\phi =-\pi \]

4.1.1.2 part b

\[ x_{2}\left ( t\right ) =\sqrt{2}e^{j\frac{\pi }{4}}\cos \left ( 3t+2\pi \right ) \] Since \(\cos \left ( 3t+2\pi \right ) =\cos \left ( 3t\right ) \), the above becomes\begin{align*} x_{2}\left ( t\right ) & =\sqrt{2}e^{j\frac{\pi }{4}}\cos \left ( 3t\right ) \\ & =\sqrt{2}\left ( \cos \frac{\pi }{4}+j\sin \frac{\pi }{4}\right ) \cos \left ( 3t\right ) \\ & =\sqrt{2}\left ( \frac{1}{2}\sqrt{2}+j\frac{1}{2}\sqrt{2}\right ) \cos \left ( 3t\right ) \\ & =\left ( 1+j\right ) \cos \left ( 3t\right ) \end{align*}

Hence the real part of \(x_{2}\left ( t\right ) \) is \[ \operatorname{Re}\left ( x_{2}\left ( t\right ) \right ) =\cos \left ( 3t\right ) \] Comparing the above to \(Ae^{-at}\cos \left ( \omega t+\phi \right ) \) shows that \[ A=1,a=0,\omega =3,\phi =0 \]

4.1.1.3 part c

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

4.1.1.4 part d

\begin{align*} x_{4}\left ( t\right ) & =je^{\left ( -2+100j\right ) t}\\ & =je^{-2t}e^{j100t} \end{align*}

But \(j=e^{j\frac{\pi }{2}}\), hence the above becomes\begin{align*} x_{4}\left ( t\right ) & =e^{j\frac{\pi }{2}}e^{-2t}e^{j100t}\\ & =e^{-2t}e^{j\left ( 100t+\frac{\pi }{2}\right ) }\\ & =e^{-2t}\left ( \cos \left ( 100t+\frac{\pi }{2}\right ) +j\sin \left ( 100t+\frac{\pi }{2}\right ) \right ) \end{align*}

Therefore \[ \operatorname{Re}\left ( x_{4}\left ( t\right ) \right ) =e^{-2t}\cos \left ( 100t+\frac{\pi }{2}\right ) \] Comparing the above to \(Ae^{-at}\cos \left ( \omega t+\phi \right ) \) shows that \[ A=1,a=2,\omega =100,\phi =\frac{\pi }{2}\]

4.1.2 Problem 1.13, Chapter 1

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

Solution

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

\(\delta \left ( t+2\right ) \) is an impulse at \(t=-2\) and \(\delta \left ( t-2\right ) \) 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\[ y\left ( t\right ) =\left \{ \begin{array} [c]{cc}0 & t<-2\\ 1 & -2<t<2\\ 0 & t>2 \end{array} \right . \] Now that \(y\left ( t\right ) \) is found, its \(E_{\infty }\) can be calculated using the definition \begin{align*} E_{\infty } & =\int _{-\infty }^{\infty }\left \vert y\left ( t\right ) \right \vert ^{2}dt\\ & =\int _{-2}^{2}1dt\\ & =\left [ t\right ] _{-2}^{2}\\ & =2+2 \end{align*}

Hence\[ E_{\infty }=4 \]

4.1.3 Problem 1.17, Chapter 1

Consider a continuous-time system with input \(x(t)\) and output \(y(t)\) related by \(y\left ( t\right ) =x\left ( \sin \left ( t\right ) \right ) \). (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 \(y\left ( -\pi \right ) =x\left ( \sin \left ( -\pi \right ) \right ) =x\left ( 0\right ) \). This shows that \(y\left ( -\pi \right ) =x\left ( 0\right ) \). 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{align*} y\left ( t\right ) & =x\left ( \sin \left ( t\right ) \right ) \\ & =a_{1}x_{1}\left ( \sin \left ( t\right ) \right ) +a_{2}x_{2}\left ( \sin \left ( t\right ) \right ) \end{align*}

Now, \(y_{1}\left ( t\right ) =x_{1}\left ( \sin t\right ) \) and \(y_{2}\left ( t\right ) =x_{2}\left ( \sin t\right ) \). Hence the above becomes\begin{align*} y\left ( t\right ) & =x\left ( \sin \left ( t\right ) \right ) \\ & =a_{1}y_{1}\left ( t\right ) +a_{2}y_{2}\left ( t\right ) \end{align*}

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) \(x\left ( t-1\right ) \). (b) \(x\left ( 2-t\right ) \) (c) \(x\left ( 2t+1\right ) \) (d) \(x\left ( 4-\frac{t}{2}\right ) \)

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\[ x\left ( t\right ) =-u\left ( t+2\right ) +r\left ( t+2\right ) -r\left ( t+1\right ) +u\left ( t+1\right ) +u\left ( t\right ) -u\left ( t-1\right ) -r\left ( t-1\right ) +r\left ( t-2\right ) \] 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

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

Figure 4.4:Part (a) plot

Figure 4.5:Code for the above

4.1.4.2 Part b

\[ x\left ( 2-t\right ) =x\left ( -\left ( t-2\right ) \right ) \] 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{align*} x\left ( 2-t\right ) & =-u\left ( \left ( 2-t\right ) +2\right ) +r\left ( \left ( 2-t\right ) +2\right ) -r\left ( \left ( 2-t\right ) +1\right ) +u\left ( \left ( 2-t\right ) +1\right ) +u\left ( 2-t\right ) -u\left ( \left ( 2-t\right ) -1\right ) -r\left ( \left ( 2-t\right ) -1\right ) +r\left ( \left ( 2-t\right ) -2\right ) \\ & =-u\left ( 4-t\right ) +r\left ( 4-t\right ) -r\left ( 3-t\right ) +u\left ( 3-t\right ) +u\left ( 2-t\right ) -u\left ( 1-t\right ) -r\left ( 1-t\right ) +r\left ( -t\right ) \end{align*}

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

\(x\left ( 2t+1\right ) =x\left ( 2\left ( t+\frac{1}{2}\right ) \right ) \). 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

\(x\left ( 4-\frac{t}{2}\right ) =x\left ( -\left ( \frac{t}{2}-4\right ) \right ) =x\left ( -\frac{1}{2}\left ( t-8\right ) \right ) \). 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) \(x\left [ n-4\right ] \) (b) \(x\left [ 3-n\right ] \) (c) \(x\left [ 3n\right ] \) (d) \(x\left [ 3n+1\right ] \)

Figure 4.11:The function \(x[n]\)

Solution

4.1.5.1 Part a

\(x\left [ n-4\right ] \) 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

\(x\left [ 3-n\right ] =x\left [ -\left ( n-3\right ) \right ] \). Hence \(x[n]\) is first reflected to obtain \(x\left [ -n\right ] \) 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

\(x\left [ 3n+1\right ] \). 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) \(\sin \left ( \frac{6}{7}\pi n+1\right ) \) (b) \(x\left [ n\right ] =\cos \left ( \frac{n}{8}-\pi \right ) \) (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 \(x\left [ n\right ] =x\left [ n+N\right ] \) for all \(n\). Fundamental period is the smallest such integer \(N\).

4.1.6.1 Part a

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

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, \(x\left [ n\right ] =\cos \left ( \frac{n}{8}-\pi \right ) \). Hence the signal is periodic if \begin{align*} x\left [ n\right ] & =x\left [ n+N\right ] \\ \cos \left ( \frac{n}{8}-\pi \right ) & =\cos \left ( \frac{n+N}{8}-\pi \right ) \\ & =\cos \left ( \left ( \frac{n}{8}-\pi \right ) +\frac{N}{8}\right ) \end{align*}

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

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{align*} x\left [ n\right ] & =x\left [ n+N\right ] \\ \cos \left ( \frac{\pi }{8}n^{2}\right ) & =\cos \left ( \frac{\pi }{8}\left ( n+N\right ) ^{2}\right ) \\ & =\cos \left ( \frac{\pi }{8}\left ( n^{2}+N^{2}+2nN\right ) \right ) \\ & =\cos \left ( \frac{\pi }{8}n^{2}+\frac{\pi }{8}\left ( N^{2}+2nN\right ) \right ) \end{align*}

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

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

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

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

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{align*} x\left [ n\right ] & =x\left [ n+N\right ] \\ \cos \left ( \frac{\pi }{4}n\right ) & =\cos \left ( \frac{\pi }{4}\left ( n+N\right ) \right ) \\ & =\cos \left ( \left ( \frac{\pi }{4}n\right ) +\frac{\pi }{4}N\right ) \end{align*}

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

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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