0
$\begingroup$

Given an input signal $$x(n)=\cos(6\pi n +\frac{\pi}{6})$$ and system $$y(n)=0.5x(n)-0.1x(n-1)$$. In this case, the coefficients of the difference equation are $a_0=1$, $b_0=0.5$, and $b_1=$. The frequency response to this system is $$H(\omega)= b_0 + b_1e^{-j\omega}= 0.5 - 0.1 \cos(\omega)+0.1 j \sin(\omega)$$

Since the input $x(n)$ is a sinusoid of the form $$x(n)=A \cos(\omega n +\theta_0)$$, we can express the system's output as

$$y(n)=A|H(\omega_0)|\cos\left(\omega_0 n+\theta_0+angle(H(\omega_0))\right)$$

I can see that $A=1$, $\omega_0=6\pi$, and $\theta_0=\frac{\pi}{6}$ in this formula. Hence, I have deduced that

$$|H(\omega_0)|= |0.5 - 0.1 \cos(6\pi)+0.1j \sin(6\pi)| = 0.4$$

Furthermore, since $H(\omega_0) = 0.4$ (a positive real number), I conjecture that $$angle(H(\omega_0))=0$$

Have I made the correct assumptions/conclusions?

Improve this question
$\endgroup$
4
  • $\begingroup$ Are you sure that $H(\omega)$ is correct? You should derive it again correctly, try to use $Y(\omega) / X(\omega)$. $\endgroup$
    – Deniz
    Commented Oct 8, 2012 at 15:21
  • $\begingroup$ @DenizAkyildiz: The frequency response of a difference equation is $\frac{\sum_{m=0}^M b_me^{-j\omega m}}{1 + \sum_{l=0}^N a_le^{-j\omega l}}$? Am I mistaken? $\endgroup$
    – Paul
    Commented Oct 8, 2012 at 15:30
  • 2
    $\begingroup$ Try to put in this way, $$Y(\omega) = 0.5 X(\omega) - 0.1 e^{- j \omega} X(\omega)$$ Then, $$H(\omega) = \frac{Y(\omega)}{X(\omega)} = 0.5 - 0.1 e^{- j \omega}$$ I don't use such formulas. If I remember correctly, the subscript of sum in the denominator should be like this: $\sum_{n=1}^N a_n \ldots$. Then, you should divide to 1. $\endgroup$
    – Deniz
    Commented Oct 8, 2012 at 15:33
  • $\begingroup$ Oh... you're absolutely right! My mistake. I'll edit the question to reflect this. $\endgroup$
    – Paul
    Commented Oct 8, 2012 at 15:38

1 Answer 1

Reset to default
1
$\begingroup$

Yes, you are correct. The system input is rather strange, though, since $cos(6\pi n + \frac{\pi}{6})$ is equal to $cos(\frac{\pi}{6})$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Actually in my case, n is not an integer, but a set of 100 points equispaced from 0 to 2*pi. Is there a better notation or convention to denote that the values in the sequence are not integers? $\endgroup$
    – Paul
    Commented Oct 14, 2012 at 20:50
  • $\begingroup$ Are you sure, because that makes no sense. First, $n$ is almost always, by convention, used to denote integers. Second, the system response ($y(n) = 0.5x(n)−0.1x(n−1)$) is clearly using $n$ as a vector of time integers. If it really goes from $0$ to $2\pi$ then that makes no sense. $\endgroup$
    – Jim Clay
    Commented Oct 14, 2012 at 21:50
  • $\begingroup$ It's the notation used in my textbook :) $\endgroup$
    – Paul
    Commented Oct 15, 2012 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.