Read the discussion on discussion page using the discussion page tab. (Aung 11:13pm on 07/18/2008) --The reasoning for (a) (d) (g) are wrong. Think again. (Aung 11:11 pm on 07/18/08)

(a) The FT of $X(j\omega)$ of a continuous-time signal x(t) is periodic

MAY BE: -


(b) The FT of $X(e^{j\omega})$ of a continuous-time signal x[n] is periodic

YES: $X(e^{j\omega})$ is always periodic with period $2\pi$


(c) If the FT of $X(e^{j\omega})$ of a discrete-time signal x[n] is given as: $X(e^{j\omega}) = 3 + 3cos(3\omega)$, then the signal x[n] is periodic

NO: The inverse transform of this signal is a set of delta functions that are not periodic.


(d) If the FT of $X(j\omega)$ of a continuous-time signal x(t) consists of only impulses, then x(t) is periodic

MAY BE: -


(e) Lets denote $X(j\omega)$ the FT of a continuous-time non-zero signal x(t). If x(t) is an odd signal, then: $\int_{-\infty}^{\infty} X(j\omega) d\omega = 0$

YES: this equation is the same as $\int_{-\infty}^{\infty} X(j\omega) e^{-j\omega_0t}d\omega = 0$ where t = 0.
From this we can conclude that x(0) = 0, which always holds true for odd signals.


(f) Lets denote $X(j\omega)$ the FT of a continuous-time non-zero signal x(t). If x(t) is an odd signal, then: $\int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0$

NO: using parseval's relation, we see that: $\int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega = 0 = 2\pi \int_{-\infty}^\infty |x(t)|^2 dt$
The integral of the magnitude squared will always be positive for an odd signal.


(g) Lets denote $X(e^{j0})$ the FT of a DT signal x[n]. If $X(e^{j0})$ = 0, then x[n] = 0.

MAY BE: $X(e^{j0})$ is simply $X(e^{j\omega})$ evaluated at $\omega = 0$.
This only tells you that summation of x[n] over all n is 0, not the entire signal x[n] = 0.


(h) If the FT ($X(e^{j\omega})$) of a discrete-time signal x[n] is given as : $X(e^{j\omega}) = e^{-j10\omega}/(22.30e^{-j5\omega} + 11.15)$ then the signal x[n] is real.

MAY BE: x[n] is real only if the properties of conjugate symmetry


for real signals hold for this transform.

(i) Let x(t) be a continuous time real-valued signal for which $X(j\omega)$ = 0 when $|\omega| > \omega_M$ where $\omega_M$ is a real and positive number. Denote the modulated signal y(t) = x(t)c(t) where c(t) = $cos(\omega_ct)$ and $\omega_c$ is a real, positive number. If $\omega_c$ is greater than $2\omega_M$, x(t) can be recovered from y(t).

YES: Taking the FT of c(t) we get delta functions at $\omega_c$ and $-\omega_c$.
When convolved with the FT of the input signal $X(j\omega)$, the function $X(j\omega)$ gets shifted to


$\omega_c$ and $-\omega_c$ with ranges $(-\omega_c-\omega_M)$ to $(-\omega_c+\omega_M)$ and $(\omega_c-\omega_M)$ to $(\omega_c+\omega_M)$.

Therefore $(\omega_c-\omega_M) > (-\omega_c+\omega_M)$ must hold for there to be no
overlapping. This is equivalent to $2\omega_c > 2\omega_M => \omega_c > \omega_M$. Since $\omega_c > 2\omega_M$,


there is no overlapping and x(t) can be recovered.

(j) Let x(t) be a continuous time real-valued signal for which $X(j\omega)$ = 0 when $|\omega| > 40\pi$. Denote the modulated signal y(t) = x(t)c(t) where c(t) = $e^{j\omega_ct}$ and $\omega_c$ is a real, positive number. There is a constraint of $\omega_c$ to guarantee that x(t) can be recovered from y(t).

NO: The FT of c(t) is just a shifted delta function, which will simply shift the
input signal x(t) so there is no chance of overlapping.


## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett