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Homework 2 Solutions, ECE301 Spring 2011 Prof. Boutin
Question 1
a)
$ E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} \left|e^{-t}u(t)\right|^2dt = \lim_{T \rightarrow \infty} \int_{0}^{T} e^{-2t}dt = \lim_{T \rightarrow \infty} -\frac{1}{2}\left[e^{-2T}-e^0\right]=\frac{1}{2} $
$ P_\infty = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} \left|e^{-t}u(t)\right|^2dt = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{0}^{T} e^{-2t}dt = \lim_{T \rightarrow \infty} -\frac{1}{4T}\left[e^{-2T}-e^0\right] = \lim_{T \rightarrow \infty} \frac{1-e^{-2T}}{4T}=0 $
Since the signal has finite energy, then we expect that it has zero average power.
b)
$ E_\infty = \lim_{T \rightarrow \infty} \int_{-T}^{T} \left|e^{jt}u(t)\right|^2dt = \lim_{T \rightarrow \infty} \int_{0}^{T} dt = \lim_{T \rightarrow \infty} T = \infty $
$ P_\infty = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} \left|e^{jt}u(t)\right|^2dt = \lim_{T \rightarrow \infty} \frac{1}{2T} \int_{0}^{T} dt = \lim_{T \rightarrow \infty} \frac{T}{2T} = \frac{1}{2} $
Since the signal has infinite energy, then we expect that it has average power that is greater than zero.
c)
$ E_\infty = \lim_{N \rightarrow \infty} \sum_{n=-N}^{N} \left|\frac{1}{3}u[n]\right|^2 = \lim_{N \rightarrow \infty} \sum_{n=0}^{N} \frac{1}{9} = \lim_{N \rightarrow \infty} \frac{1}{9}(N+1) = \infty $
$ P_\infty = \lim_{N \rightarrow \infty} \frac{1}{2N+1}\sum_{n=-N}^{N} \left|\frac{1}{3}u[n]\right|^2 = \lim_{N \rightarrow \infty} \frac{1}{2N+1} \sum_{n=0}^{N} \frac{1}{9} = \lim_{N \rightarrow \infty} \frac{1}{9} \cdot \frac{N+1}{2N+1} = \frac{1}{9} \cdot \frac{1}{2} = \frac{1}{18} $
Question 2
a)
$ x[n+N] = e^{j\frac{3}{5}\pi(n+N-1/2)} = e^{j\frac{3}{5}\pi N} \cdot e^{j\frac{3}{5}\pi(n-1/2)} $
For x[n + N] to be equal to x[n], $ e^{j\frac{3}{5}\pi N} $ should be equal to one.
This implies that 3πN / 5 = 2πK, where k is an integer, or N = 10k / 3. Now, the smallest positive integer N that is not zero is 10. Then the fundamental period of this signal is 10.
b)
$ x(t)=\cos^2 t = \frac{1}{2}+\frac{1}{2}\cos(2t) $
$ x(t+T)= \frac{1}{2}+\frac{1}{2}\cos(2t+2T) $
x(t + T) = x(t) for T = πk, where k is an integer. Now, the smallest positive nonzero T is π, and hence the fundamental period is π.
c)
$ x[n]=\cos^2 n = \frac{1}{2}+\frac{1}{2}\cos[2n] $
$ x[n+N]= \frac{1}{2}+\frac{1}{2}\cos[2n+2N] $
x[n + N] = x[n] for N = πk, where k is an integer. Since x[n] is a discrete-time signal and N is a multiple of π, i.e. any non-zero N is not an interger, then we can say that the signal is not periodic.
d)
$ \begin{align} x[n+N] &= 1 + e^{j\frac{4\pi}{7}(n+N)}-e^{j\frac{2\pi}{5}(n+N)} \\ &= 1+e^{j\frac{4\pi}{7}N}\cdot e^{j\frac{4\pi}{7}n} - e^{j\frac{2\pi}{5}N} \cdot e^{j\frac{2\pi}{5} n} \\ \end{align} $
We can see that for N = 35k, where k is an integer, x[n + N] = x[n]. Then the fundamental frequency is 35.
Note that we can find the fundamental frequency of this signal directly by knowing that the fundamental period of the sum of periodic signals is the least common multiple of the periods of the individual signals. For this specific signal, the first term has a fundamental period of 1, the second term has a fundamental period of 7, and the third term has a fundamental period of 5. Thus the fundamental period of the sum of these terms or signals is the least common multiple of 1, 7, and 5 which is 35.
Note also that the fundamental period of a complex exponential of the form $ e^{j\frac{2\pi}{N}n} $ is N.
e) If we let $ f(t)=\frac{1}{1+t^2} $, then x(t) can be written in the form of $ x(t) = \sum_{k=-\infty}^{\infty}f(t-7k) $.
Then the fundamental period is 7.
How does the above show that the fundamental period is 7?
It shows that the period is 7 because x(t) is a sum of shifted copies of f(t). Observe that the copies are shifted by 7 units with respect to each other. So x(t+7) can be obtained by moving each of the shifted copies of f(t) by 7 units towards the left. In effect, one shifted copy is put in place of the one to the left of it, and so the summation remains the same as before. You should be able to prove this mathematically from your lecture notes. (See this page for more details.)
Now to show that 7 is the FUNDAMENTAL period is a more complicated story... -pm
Question 3
$ x_e[n]=\frac{x[n]+x[-n]}{2} $
$ x_o[n]=\frac{x[n]-x[-n]}{2} $
Now,
$ \begin{align} \sum_{n=-\infty}^{\infty}x_e^2[n]+\sum_{n=-\infty}^{\infty}x_o^2[n] &= \sum_{n=-\infty}^{\infty}\left(x_e^2[n]+x_o^2[n]\right) \\ &= \sum_{n=-\infty}^{\infty}\frac{x^2[n]+2x[n]x[-n]+x^2[-n] + x^2[n]-2x[n]x[-n]+x^2[-n]}{4} \\ &= \sum_{n=-\infty}^{\infty}\frac{x^2[n]+x^2[-n]}{2} \\ &= \frac{1}{2} \sum_{n=-\infty}^{\infty}x^2[n] + \frac{1}{2} \sum_{n=-\infty}^{\infty}x^2[-n] \\ &= \frac{1}{2} \sum_{n=-\infty}^{\infty}x^2[n] + \frac{1}{2} \sum_{m=-\infty}^{\infty}x^2[m] \\ &= \sum_{n=-\infty}^{\infty}x^2[n]. \end{align} $
where we have used the change of variable m=-n.
Question 4
We choose A440 as the single note that we will use to write out function z(t).
Call this single note a(t), then a(t) = s'i'n(2πfAt)[u(t) − u(t − 1)], where fA = 440 Hz.
Here are the 12 notes of "Smoke on the Water" melody as a function of a(t) listed in order of time:
$ x_1(t)=a\left(2^{-\frac{1}{6}}t\right)\left[u(t)-u\left(t-\frac{60}{BPM}\right)\right] $
$ x_2(t)=a\left[2^{\frac{1}{12}}\left(t-\frac{60}{BPM}\right)\right]\left[u(t-\frac{60}{BPM})-u\left(t-\frac{120}{BPM}\right)\right] $
$ x_3(t)=a\left[2^{\frac{1}{4}}\left(t-\frac{120}{BPM}\right)\right]\left[u(t-\frac{120}{BPM})-u\left(t-\frac{210}{BPM}\right)\right] $
$ x_4(t)=a\left[2^{-\frac{1}{6}}\left(t-\frac{210}{BPM}\right)\right]\left[u(t-\frac{210}{BPM})-u\left(t-\frac{270}{BPM}\right)\right] $
$ x_5(t)=a\left[2^{\frac{1}{12}}\left(t-\frac{270}{BPM}\right)\right]\left[u(t-\frac{270}{BPM})-u\left(t-\frac{330}{BPM}\right)\right] $
$ x_6(t)=a\left[2^{\frac{1}{3}}\left(t-\frac{330}{BPM}\right)\right]\left[u(t-\frac{330}{BPM})-u\left(t-\frac{360}{BPM}\right)\right] $
$ x_7(t)=a\left[2^{\frac{1}{4}}\left(t-\frac{360}{BPM}\right)\right]\left[u(t-\frac{360}{BPM})-u\left(t-\frac{480}{BPM}\right)\right] $
$ x_8(t)=a\left[2^{\frac{-1}{6}}\left(t-\frac{480}{BPM}\right)\right]\left[u(t-\frac{480}{BPM})-u\left(t-\frac{540}{BPM}\right)\right] $
$ x_9(t)=a\left[2^{\frac{1}{12}}\left(t-\frac{540}{BPM}\right)\right]\left[u(t-\frac{540}{BPM})-u\left(t-\frac{600}{BPM}\right)\right] $
$ x_{10}(t)=a\left[2^{\frac{1}{4}}\left(t-\frac{600}{BPM}\right)\right]\left[u(t-\frac{600}{BPM})-u\left(t-\frac{690}{BPM}\right)\right] $
$ x_{11}(t)=a\left[2^{\frac{1}{12}}\left(t-\frac{690}{BPM}\right)\right]\left[u(t-\frac{690}{BPM})-u\left(t-\frac{750}{BPM}\right)\right] $
$ x_{12}(t)=a\left[2^{-\frac{1}{6}}\left(t-\frac{750}{BPM}\right)\right]\left[u(t-\frac{750}{BPM})-u\left(t-\frac{810}{BPM}\right)\right] $
Finally, we can write z(t) as the sum of the above signals: $ z(t)=x_1(t)+x_2(t)+\dots+x_{12}(t) $.
Question 5
Denote the outputs of system 1 and system 2 by v(t) and z(t), respectively. Then, we have:
v(t) = x(3t + 7),
z(t) = v(5t − 1) = x(3(5t − 1) + 7) = x(15t + 4), and
y(t) = z( − t) = x(15( − t) + 4) = x( − 15t + 4).
Hence, the output of the cascade is y(t) = x( − 15t + 4).
