Suppose the input to an FIR system is

$x[n]=e^{j(2\pi n/7 -\pi/2)}$

Define a new signal $y[n] = x[n]-x[n-1]$ . The signal $y[n]$ can be expressed in the form

$y[n] = Ae^{j( 2 \pi f n + \phi)}$

Find $A$, $\phi$, and $f$.

I really would just like confirmation that I am correct.
I calculated $A = 0.8678$

$\phi = -1.7952$

$f = 0.2244$


1 Answer 1


Unfortunately, apart from the amplitude $A$, I can't confirm that you're correct. The first thing that you can see without doing any calculations is that $f$ must be the same for input and output, because the system is linear and time-invariant, so the frequency of the complex exponential remains unchanged. Consequently, $f=1/7$. The output signal can be written as

$$\begin{align}y[n]&=e^{j(2\pi n/7-\pi/2)}\cdot\left(1-e^{-j2\pi/7}\right)\\&=e^{j(2\pi n/7-\pi/2)}\cdot e^{-j\pi /7}\left(e^{j\pi /7}-e^{-j\pi /7}\right)\\&=e^{j(2\pi n/7-\pi/2)}\cdot e^{-j\pi /7}\cdot 2j\sin(\pi/7)\\ &=2\sin(\pi/7)\cdot e^{j(2\pi n/7-\pi/7)} \end{align}$$

So you have

$$\begin{align}A&=2\sin(\pi/7)=0.8678\\ f&=1/7=0.1429\\ \phi&=-\pi/7=-0.4488 \end{align}$$


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