Discussion For Homework1

If there are any questions/comments about homework 1 that you would like to share
with the class, please post them here.


Comment: In problem 4, the graph should not have $\phi_{NN} (f)$ labeled on the vertical axis, but instead $\Phi_{NN} (f)$. A small error.


Question: In problems 6 and 7, what does it mean when it says "with respect to a carrier frequency f0 of x Hz"

Answer[Krishna]: With respect to a carrier frequency of $f_0$ Hz means that the passband signal should be written as

(1)
\begin{align} s(t) = x(t) \ \cos(2 \pi f_0 t) - y(t) \ \sin(2 \pi f_0 t) \end{align}

Notice that there is not a unique $f_0$ for which we can write the signal $s(t)$ like this. The last problem in the homework or may be even the last two problems show that this is in fact true.

Comment[student]: There will always be some minor difference between the carrier frequency that is used to transmit and the frequency at the receiver oscillator. I think the questions 6 and 7 are trying explain the affect of that difference on the excess phase part of base band signal.
We can always get I and Q of a signal, by two ways:
1. Either by expressing the given signal in the form (I cos$\omega_{c}t$ - Q sin$\omega_{c}t$) or
2. By multiplying the given signal with the carrier frequency and low pass filter.

Second method is suitable for band limited signals.

Question: Can you provide an example of method 2 as mentioned above? I am familiar with method 1, but not so much with method 2. Thanks.

Question: In problem 6 of the homework, the given signal S(t) is the base band signal …. isn't it? Also, it is a sinc function in the time domain which is symmetric around the y-axis in the frequency and thus shouldn't have any quadrature component without any conditions. Is that true?

Answer[Krishna]: Let us not think of baseband or passband signals for this problem. Given the signal $s(t)$, I want you to find an $f_0$ for which we can write this as

(2)
\begin{align} s(t) = x(t) \ \cos(2 \pi f_0 t) - y(t) \ \sin(2 \pi f_0 t) \end{align}

such that $y(t) = 0$.


Comment: Consider a given passpand signal $s(t)$. We write it with respect to carrier frequency $f_0$ as

(3)
\begin{align} s(t) = x(t) \ \cos(2 \pi f_0 t) - y(t) \ \sin(2 \pi f_0 t). \end{align}

Note that we can assign different values to $f_0$, which result in different baseband signals $x(t)+j y(t)$. But not all $f_0$'s meat the definition of ''narrowband signals". For narrowband signals, the frequency components of the obtained $x(t)$ and $y(t)$, which are within the baseband, must be much lower (at least not higher) than the seleted $f_0$.

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