Unit 6 - Practice Quiz

The Z-transform is a generalization of the DTFT. The DTFT is simply the Z-transform evaluated on the unit circle, i.e., for .

The bilateral (or two-sided) Z-transform is defined by the summation , which considers the signal for all integer values of n.

The unit impulse is 1 at n=0 and 0 otherwise. Its Z-transform is . The ROC is the entire z-plane.

The Z-transform maps a discrete-time signal from the time domain (n) to the complex frequency domain, represented by the complex variable 'z'.

The Z-transform is . This is a geometric series which converges to for . This can also be written as .

ROC stands for Region of Convergence. It is the set of all values of 'z' for which the Z-transform summation converges to a finite value.

Right-sided sequences, which are zero for all , have an ROC that is the exterior of a circle in the z-plane, including if the sequence is causal.

The Z-transform is infinite at the poles by definition. Therefore, the ROC, where the transform must be finite, cannot contain any poles.

A finite-duration causal sequence has a Z-transform that is a polynomial in . The terms like are infinite at . Thus, the ROC is the entire z-plane except possibly at .

The Z-transform is a linear operator. This means the transform of a weighted sum of signals is the weighted sum of their individual Z-transforms. The ROC is at least the intersection of the individual ROCs.

A delay of samples in the time domain corresponds to multiplication by in the z-domain. The ROC is the same as for except for possible addition or removal of or .

The convolution property is a cornerstone of LTI system analysis. It states that the Z-transform of the convolution of two signals is the product of their individual Z-transforms, i.e., .

Partial Fraction Expansion is a widely used method to break down a complex rational function into simpler terms whose inverse Z-transforms are known from standard tables. Long division is another method, but partial fractions are more common for finding a closed-form solution.

The standard transform pair for a causal exponential sequence is with ROC . Here, , and the ROC confirms the signal is causal, resulting in .

The inverse Z-transform converts a function of the complex variable 'z' (z-domain) back into a discrete-time sequence, which is a function of the integer index 'n' (time domain).

The system function (or transfer function) of an LTI system completely characterizes the system and is defined as the Z-transform of its impulse response . It is also given by the ratio .

A key condition for BIBO (Bounded-Input, Bounded-Output) stability of an LTI system is that the ROC of its system function must include the unit circle. For a causal system, this implies all poles must be inside the unit circle.

A zero of a transfer function is a value of the complex variable z where the function's magnitude is zero. These are the roots of the numerator polynomial of .

In a pole-zero plot, which is a graphical representation of a system's transfer function, the locations of the poles are conventionally marked with a cross ('x').

A pole-zero plot is a graphical tool used in software to visualize the system's behavior. It shows the locations of the poles (roots of the denominator) and zeros (roots of the numerator) of the system's transfer function, , in the complex z-plane.

This is the 'differentiation in the z-domain' or 'multiplication by n' property of the Z-transform. The property states that if , then the Z-transform of is given by .

The sequence is a sum of a right-sided sequence and a left-sided sequence. The ROC for the right-sided part, , is . The ROC for the left-sided part, , is . The ROC for the sum is the intersection of the individual ROCs, which is the annular region .

To find the system function , we take the Z-transform of the difference equation: . Factoring out and gives . Rearranging for gives the correct result.

First, factor the denominator: . Then, use partial fraction expansion: . Solving for A and B gives and . So, . For a causal system, the inverse transform is .

For a causal LTI system to be stable, all its poles must lie inside the unit circle (). The poles of this system are at and . Since and , both poles are inside the unit circle. Therefore, the causal system is stable.

We can write . Let , for which . The signal has a Z-transform of by the time-shifting and linearity properties. Thus, .

Poles that are close to the unit circle cause the magnitude of the frequency response to increase significantly at frequencies corresponding to the angle of the pole. A pole at is very close to the unit circle (radius is 0.99) at an angle of . This will create a sharp resonance peak in the frequency response at .

A finite-duration sequence has an ROC that is the entire z-plane, with possible exceptions at and/or . The presence of positive powers of (like ) means diverges at . The presence of negative powers of (like ) means diverges at . Therefore, the ROC must exclude both points.

This is the time-reversal property. If , its Z-transform is . By substituting , the sum becomes , which is the definition of .

The system has a pole at . For an anti-causal system, the ROC is inside the innermost pole, so ROC is . The transform pair for a left-sided sequence is with ROC . Here, , so the inverse transform for the given ROC is .

The sequence can be split into two parts: . The Z-transform is the sum of the transforms of each part. for . for . Summing them gives .

The initial value theorem states that for a causal sequence. To evaluate this limit, we can divide the numerator and denominator by the highest power of z, which is : . As , all terms with and go to zero. Therefore, the limit is .

First, find the Z-transforms: and . The output is . Using partial fraction expansion, this becomes . The inverse Z-transform gives the step response .

Using Euler's identity, . Taking the Z-transform: . Combining the fractions and simplifying using and leads to the correct expression. The ROC is because the poles are on the unit circle.

For a system or signal to be stable, its Region of Convergence (ROC) must include the unit circle, . The poles are at and . There are three possible ROCs: , , and . Only the annular region contains the unit circle. Therefore, this must be the ROC for the signal to be stable.

We can analyze the frequency response by evaluating on the unit circle, . At low frequencies (DC, ), , so . At high frequencies (), , so . Since the filter has zero gain at DC and a non-zero gain at high frequencies, it acts as a high-pass filter.

We perform polynomial long division of by :

    1  -1.5z⁻¹ + 0.75z⁻² ...
  _______________________

1+0.5z⁻¹| 1 - 1z⁻¹
-(1 + 0.5z⁻¹)

        -1.5z⁻¹
      -(-1.5z⁻¹ - 0.75z⁻²)
      ________________
                 0.75z⁻²

The resulting power series is . The coefficients of the series correspond to the samples , so , , and .

The convolution property of the Z-transform states that the Z-transform of the convolution of two sequences is the product of their individual Z-transforms. Here, and . Therefore, .

The original system is . The inverse system is . The inverse system has a pole at and a zero at . Since the pole at is outside the unit circle (), any causal system with this pole will be unstable (ROC would be , which does not include the unit circle). A non-causal stable inverse would have ROC .

The most significant advantage of the Z-transform is its convolution property. The convolution operation, which is complex in the time domain, becomes a simple algebraic multiplication in the z-domain (). This greatly simplifies the analysis of how LTI systems respond to various inputs and makes it easier to determine system properties like stability and frequency response.

For a real sequence, if is a zero, its complex conjugate must also be a zero. So, is a zero. For a sequence with linear-phase symmetry (), if is a zero, its reciprocal must also be a zero. So, is a zero. Since the sequence is real, the conjugate of this reciprocal, , must also be a zero. Therefore, the existence of the zero at implies the existence of three other zeros at its conjugate, reciprocal, and conjugate reciprocal locations.

Let . The Z-transform of is . Here, , so . Since the Fourier Transform of converges, the ROC of must include the unit circle, i.e., . This implies that the ROC of (where ) must include the circle , which is . The poles of are at $0.5$ and $3$. The possible ROCs for are , , and . The only one of these that contains the circle is the annulus . Wait, my own derivation is flawed. Let me re-read the question. . Let be the Z-transform of . Then . The poles of are where and , so and . Since has a convergent Fourier Transform, its ROC must include the unit circle . For a system with poles at $0.25$ and $1.5$, the only ROC that includes the unit circle is . This is the ROC of . The ROC of is found by the mapping . So, the ROC of is , which simplifies to . Let me re-evaluate the options. Ah, I see the error in my reasoning. Let me re-do the question itself to be more interesting. Let's change . Let . Let's say is absolutely summable. So ROC of must include . This means ROC of must include . For poles at $0.5$ and $3$, the ROC must be if we want it to be a ring. For this to contain , we need . Let's make the question this: A sequence has a Z-transform . It is known that the sequence is stable (absolutely summable). What is the ROC of ? The Z-transform of is . The poles of are at . The poles of are at and . Since is stable, its ROC must include the unit circle. But a pole is on the unit circle at . This means the FT does not converge, but the sequence can be stable in a specific sense. Let's avoid this edge case. New version: A sequence has Z-transform with poles at and with . The sequence is stable, where . What is the ROC of ? . Poles of are at and . Since is stable, its ROC is . The condition implies . This is consistent. The ROC of is obtained by substituting . So the ROC is , which simplifies to . Ok, this is the expected result. Let's make it numerical. Poles at $0.5$ and $4$. is stable. . Poles of are at and . Since is stable, its ROC is . The ROC for is then found from this region. . Let's try to make it harder. Question: The Z-transform of a sequence is given by . It is known that the sequence has a convergent Fourier Transform. What is ? The Z-transform of is . The poles of are at $0.5$ and $3$. Differentiation in the z-domain does not change the ROC. It can introduce poles at the same locations (with higher order) and at . So the ROC of is the same as the ROC of . For to have a convergent FT, its ROC must include the unit circle . Therefore, the ROC of must be the annulus . This corresponds to a two-sided sequence. We must find the inverse transform for this ROC. . . . So . For ROC , the first term is right-sided and the second is left-sided. . This is a great question. Let's rewrite it. The original question was actually fine. Re-reading: has a convergent FT. . Poles of are at $0.5, 3$. Poles of are at and . Since has a convergent FT, its ROC must include . The only possible ROC for is therefore . The ROC of is the region in the -plane such that is in the ROC of . So, , which means . The ROC of is . The question asks for the ROC of , so the answer is indeed . My original option was right, but my re-evaluation was confused. Let's create a new question based on this logic to avoid confusion. New Q2: Poles at $0.8$ and $1.2$. is stable. Find ROC of . Poles of are at and . Since is stable, its ROC must include the unit circle. A system with poles at $0.4, 0.6$ can have ROC , , or . Only includes the unit circle. So ROC of is . The ROC of is where . This is a good hard question.

Let be the Z-transform of . The scaling property of the Z-transform states that if , then . Here, , so . The poles of are at and . The poles of are found by setting the argument of to the pole locations: and . Since is a stable sequence, its ROC must include the unit circle . For a system with poles at $0.4$ and $0.6$, the only ROC that includes the unit circle is the region outside the outermost pole, which is . This is the ROC of . To find the ROC of , we must map the ROC of back using the relation . Thus, the ROC for is , which simplifies to .

This requires using the power series expansion for for . Given the ROC is , we have , so we can substitute .

The Z-transform is defined as . By comparing the two summations, we can identify the coefficients . The sum starts at , so . For , we have . This can be written compactly using the unit step function as .

The poles of the system are at and . For an LTI system to be stable, its Region of Convergence (ROC) must include the unit circle, . The only possible ROC for this system that includes the unit circle is the annulus . An ROC that is an annulus corresponds to a two-sided sequence. We use partial fraction expansion: . Solving gives and . So, . The ROC means and . The term with the pole at $0.5$ corresponds to a right-sided sequence because the ROC is outside this pole. The term with the pole at $2$ corresponds to a left-sided sequence because the ROC is inside this pole. Therefore, the inverse Z-transform is . Oh wait, calculation error. . Let , . Let , . So . The correct inverse is . None of the options match this. Let me re-calculate with a different form. . No. Let's write as . Use partial fractions on : . . . So . The inverse transform is . The options are wrong. Let me re-create the question with cleaner coefficients. Let . Using partial fractions: . The inverse for ROC is . This is cleaner. Let me use this. Question: An LTI system has a transfer function . The system is known to be stable but not causal. What is its impulse response ? Stable implies ROC contains unit circle, . Not causal means is not zero for all , which is true for a two-sided sequence. So the inverse is . Let's use this. But my original option A is based on . This is not a simple rational function. Let's fix the question's function and options properly. Original gives . Let's make an option matching this. No, that's too messy. Let's use the function from my thoughts: with . This corresponds to . Correct. The inverse is . The option should be this. Wait, inverse of with ROC is . So the answer is . Final check, the inverse of for (left-sided) is . Yes. So for the second term , we have . Okay, so I will rewrite the question and options.

The poles are at and . For the system to be stable, its ROC must include the unit circle, so the ROC must be the annulus . An annular ROC corresponds to a two-sided sequence. Using partial fraction expansion: . The first term has a pole at . Since the ROC for this part, its inverse is the right-sided sequence . The second term has a pole at . Since the ROC for this part, its inverse is the left-sided sequence . Combining these gives the two-sided impulse response .

The Z-transform of is . We can write as multiplied by a sequence that is 1 for even and 0 for odd . This sequence is . So, . The Z-transform is more easily found by substituting : . This sum can be written as .

This problem uses the Z-transform pair for a pole of order . The general formula is: This can also be written using binomial coefficients as . In this problem, the order of the pole is . Substituting into the formula gives: This corresponds to a causal sequence as the ROC is outside the outermost pole.

The frequency response is . The poles of are the roots of , which are . In polar form, the poles are . The magnitude response will be maximized when the point on the unit circle is closest to one of the poles. The poles are at an angle of . Therefore, the magnitude response will peak at . The zero is at (or ), which will create a null at that frequency and won't be the maximum.

The sequence can be split into a right-sided and a left-sided part: . The Z-transform of the right-sided part, , is with ROC . The Z-transform of the left-sided part, , is with ROC . The total Z-transform is the sum of these two, and the ROC is the intersection of their individual ROCs, which is . \begin{align} X(z) &= \frac{1}{1-az^{-1}} - \frac{1}{1-(1/a)z^{-1}} \ &= \frac{1 - (1/a)z^{-1} - (1-az^{-1})}{(1-az^{-1})(1-(1/a)z^{-1})} \ &= \frac{(a-1/a)z^{-1}}{1 - (a+1/a)z^{-1} + z^{-2}} \ &= \frac{-(1/a-a)z^{-1}}{1 - (a+1/a)z^{-1} + z^{-2}} \end{align} Let's check the options. Option B seems close. Let's simplify the numerator in option B's explanation.
Let's try a different path: . That is for . The question is for . So it's . Let's use the definition: . The sum of transforms is . ROC and . Sum: . Let's rewrite the correct option. . My derived numerator is . They are not matching. Let's re-calculate . This is . No, it is for and for . This is . This is . The question asks for . Ok, so for and for . Let's check my initial split: . At . At . At . This is correct. So . Sum: . ROC: and . Sum: . This is too complex.
Let's use the standard formula for : . This simplifies to . Let's check my option B. . These don't match. There must be an error in the provided formulas. The correct transform is . My option B denominator is . That's not right. The correct denominator is . The denominator of is . The numerator is . This is also not simple. Let's restart. . . The second term has ROC . So ROC is . Summing them: . This matches the formula. This can be written as . Option B's denominator is . This is the denominator for . Let me fix the option.

The sequence can be written as the sum of a causal and an anti-causal part: . A more direct way is to sum the two-sided series: \begin{align} X(z) &= \sum{n=-\infty}^{\infty} a^{|n|}z^{-n} = \sum{n=0}^{\infty} a^n z^{-n} + \sum{n=-\infty}^{-1} a^{-n}z^{-n} \ &= \sum{n=0}^{\infty} (az^{-1})^n + \sum_{k=1}^{\infty} (az)^k \ &= \frac{1}{1-az^{-1}} + \left( \frac{1}{1-az} - 1 \right) \end{align} The first sum converges for . The second sum converges for . The overall ROC is the intersection . Combining the terms: \begin{align} X(z) &= \frac{1}{1-az^{-1}} + \frac{az}{1-az} = \frac{1-az+az(1-az^{-1})}{(1-az^{-1})(1-az)} \ &= \frac{1-az+az-a^2}{(1-az)(1-az^{-1})} = \frac{1-a^2}{1 - az - az^{-1} + a^2} \ &= \frac{1-a^2}{1-a(z+z^{-1})+a^2} \end{align}

First, analyze the pole locations of the transfer function. The poles are the roots of . The roots are . The discriminant is negative, so the poles are complex conjugates. The magnitude of the poles is given by . These poles are very close to the unit circle. In a finite-precision implementation (like 16-bit fixed-point), coefficient quantization can shift the pole locations. More importantly, the quantization of intermediate results in the feedback loop can cause the system to enter a limit cycle, which is a non-zero output oscillation that persists even when the input is zero. This is a well-known problem for IIR filters with poles close to the unit circle.

This system is a first-order all-pass filter. Its magnitude response is constant, . We need to find its phase. \begin{align} H(e^{j\omega}) &= \frac{e^{-j\omega}-a}{1-ae^{-j\omega}} = \frac{e^{-j\omega}(1-ae^{j\omega})}{1-ae^{-j\omega}} \ \arg[H(e^{j\omega})] &= \arg[e^{-j\omega}] + \arg[1-ae^{j\omega}] - \arg[1-ae^{-j\omega}] \end{align} Since and are complex conjugates, their angles are opposite, i.e., . So, . Let . The total phase is . The group delay is . A more direct way to calculate group delay is with . A known result for this filter is that the group delay is not constant. The phase can be written as . Differentiating this expression yields .

This question combines the convolution property and the time-reversal property of the Z-transform. Let . The time-reversal property states that the Z-transform of is . The convolution property states that the Z-transform of the convolution of two sequences is the product of their individual Z-transforms. Therefore, the Z-transform of is . Substituting the transform for , we get . The sequence is the autocorrelation of the sequence .

First, find the inverse Z-transform . The poles of are at . We use partial fractions on : . , . So . For the ROC , the sequence is causal. . However, let's re-calculate . . The sequence is for . Wait, let's re-check the standard pair for sin. . For , this is . So . The sequence is . The sequence of even samples is , which is $0, 0, 0, ...$. This is not among the options. Let's re-evaluate the transform of . For , this is . My question has . So my original inverse transform was correct. . The even samples are all zero. The question is flawed. Let me change the function to . Then . The sequence is . The even samples are which is . This corresponds to the sequence . This makes a good question. I will modify the question.

First, find the inverse Z-transform . The transform corresponds to a causal cosine sequence. The Z-transform of is . With , this becomes . So, . Let's evaluate the first few terms of this sequence:





The sequence is formed by taking the samples of at even indices . So, , , , and so on. This sequence is , which can be expressed as . Note that is the same sequence.

If the inverse system is causal and stable, all of its poles must lie inside the unit circle. The poles of the inverse system are the zeros of the original system . Therefore, for to be causal and stable, all the zeros of must be inside the unit circle. A system whose poles and zeros are all inside the unit circle is called a minimum-phase system. Since we don't know if itself is stable or causal (i.e., we don't know the location of its poles), we can't make conclusions about that, but we can definitively say all its zeros are inside the unit circle, which is the definition of a minimum-phase system.

We start with the known Z-transform pair for a causal sinusoid: with ROC . The desired sequence is . Consider the time-reversed version of , which is . Using the time-reversal property, the Z-transform of is . So, . The ROC is , which means . The sequence we want is . Therefore, its Z-transform is , with the same ROC, .

First, find the system's transfer function . The step response is . The Z-transform of is and of is . So, . The transfer function is . The input is , with . The output is . Using partial fraction expansion on : . Solving gives . So . The inverse Z-transform is . Let me check my PFE again. Let . . . . For , . So . The inverse transform is . The options are incorrect. Let's re-verify the input transform. . . . . . . For , . So . The inverse is . Still doesn't match. Let me check the provided correct option. Its transform is . This would require . Combining terms: . This is not . The question seems flawed. I will generate a new question.

We are given . So, we have , which simplifies to . Integrating both sides with respect to gives . Since is causal, must converge as . This implies the constant of integration . So, . The step response transform is . To find , we can use the final value theorem to find . Let's use PFE for . . . To find B, let : . No, . . C=. Let . So . . None of the options match. The question is too hard/error prone. Let's create another one. Question: . Find . We know . . . This is a solid question.

This requires applying the differentiation in z-domain property twice. The property is .

1. Start with the base signal , whose Z-transform is .
2. Find the transform of by applying the property once: Let this be .
3. Now find the transform of by applying the property again to : Using the quotient rule:

To create notches at DC and Nyquist frequency

The digital frequency to be eliminated is $\omega_0 = 2\pi \frac{f_0