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1

If you go by the definition of FIR and IIR filters FIR filters : depend only on present and past input IIR filters : depend on present and past input as well as past outputs So if your difference equation has no terms involving past outputs then it is a FIR filter


1

If you have a filter impulse response represented by H(z), replacing z with z^2 is equivalent to inserting a zero-valued sample in between each original impulse response sample. Such an operation compresses the frequency response of your original filter by a factor of two.


1

It is just like ordinary functions manipulation. You have to remember that z and the frequency are linked with $z = exp(j \omega \cdot T)$ $H(z^2)$ you are doing something equivalent to $H(2 \omega)$ on the continuous frequency domain, i.e. you are narrowing the filter band, and thus stretching its time response. On the second example, you have to remember ...


2

It's probably more straightforward to first compute the total transfer function in terms of $H(z)$, then plug in the actual expression for $H(z)$, and from that write down the difference equation: $$Y(z)=X(z)+\alpha z^{-M}H(z)Y(z)\tag{1}$$ From $(1)$ we can derive the total transfer function: $$\frac{Y(z)}{X(z)}=\frac{1}{1-\alpha z^{-M}H(z)}\tag{2}$$ ...


1

When discussing the z-transform, it is important to note that there are two different frequencies that are innately involved. In many text books, the two frequencies are denoted by $\omega$ (discrete-time) and $\Omega$ (continuous-time). We'll use this convention. The z-transform uses the "discrete frequency" $\omega$ which is derived from the &...


0

In more general sense the z in Z-transform is z=r*e^jw . where r is the radius and w is the frequency. The Fourier transform is evaluated on the unit circle of the z-plane diagram where r=1.


4

Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no $z$ blocks in there to predict the future, just $z^{-1}$ block to react to the past. Also by your method of finding the transfer function, the system is causal -- with a $3^{rd}$ order numerator and ...


6

Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial conditions are given), and in general you can't see it from the transfer function either (if no region of convergence is given). The only case for which the expression ...


2

The system is causal, provided that the recursion is forward; i.e., it's recursed for increasing $k$. Seeing that you are confused about causality tests, let me elaborate on it. Let's put the definition of causality from Oppenheim's Signals & Systems book : A system is causal if the output at any time depends only on values of the input at the present ...


1

I got misled a bit in my initial reasoning. So, back to it. If we have Equation $(1)$ here below \begin{align} H\left(e^{j0.5\pi}\right) &= \frac{1 - e^{-j0.5\pi}}{1 - 0.25e^{-j\pi}}\\ &= \frac{1 - e^{-j\frac{\pi}{2}}}{1 - \frac 14e^{-j\pi}}\tag{1}\\ \end{align} And we know that $e^{j\theta} = \cos(\theta) + j\sin(\theta)$. With this we have that \...


1

By plotting the unit circle you will find that $1-e^{-j\pi/2} = \sqrt{2}e^{j\pi/4}$.


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