Unit 1 - Practice Quiz

A continuous-time signal, denoted as , is a function where the independent variable is continuous. It has a value for all points in a given time interval.

In signal processing, square brackets and an integer index like are standard notation for a discrete-time signal, which is defined only at specific, discrete points in time.

By definition, an energy signal is a signal with finite total energy, i.e., . Such signals must approach zero as .

By definition, a power signal is a signal with finite and non-zero average power, i.e., . Periodic signals are a common example of power signals.

The transformation for shifts the signal to the right along the time axis, which corresponds to a time delay of units.

Replacing the time variable with reflects the signal about the vertical axis (). This operation is called time reversal or reflection.

This is the mathematical definition of a periodic signal. The smallest positive value of for which this holds is called the fundamental period.

The fundamental period is calculated as . For this signal, the angular frequency is . Therefore, seconds.

An even signal exhibits symmetry with respect to the y-axis, meaning its value at time is the same as its value at time . The cosine function is a classic example.

An odd signal exhibits anti-symmetry with respect to the origin, meaning its value at time is the negative of its value at time . The sine function is a classic example.

This is the general form of a complex exponential signal, which is fundamental in signal analysis. It is related to sinusoidal signals through Euler's formula.

When the coefficient in the exponent is negative (since , is negative), the value of the function decreases as time increases, resulting in an exponential decay.

The unit step function, , is a basic signal that is 0 for negative time and 'steps' up to 1 at , remaining 1 for all non-negative time.

The discrete-time unit impulse, or unit sample, is a signal that has a value of 1 only at the origin () and is zero for all other integer values of .

The unit impulse function is the derivative of the unit step function with respect to time, i.e., . Conversely, the step is the integral of the impulse.

This operation scales the amplitude (vertical axis) of the signal at every point in time by the constant factor A. It makes the signal 'taller' or 'shorter'.

MATLAB (Matrix Laboratory) is a powerful tool specifically designed for matrix manipulations, plotting of functions, and implementation of algorithms, making it ideal for simulating signal operations.

When the time variable is multiplied by a constant where , the signal is stretched or expanded in time by a factor of . In this case, it is expanded by a factor of .

Periodic signals like have infinite total energy but a finite average power (for , the average power is 1/2). Therefore, it is classified as a power signal.

A signal is even if . Since , the cosine function is a classic example of an even signal.

Let the argument of the function be . The signal is non-zero for . We substitute to find the corresponding interval for : \n \n Subtract 4 from all parts: . \n Divide by -2 and reverse the inequalities: . \n Therefore, the signal is non-zero for .

The energy of a discrete-time signal is given by . \n For , the sum is from to . \n . \n This is an infinite geometric series with and . The sum is given by .

The first component, , has a period . \n The second component, , has a period . \n The fundamental period of the sum is the least common multiple (LCM) of and . \n . To find the LCM of rational numbers, we can write for the smallest integers . \n . The smallest integer solution is and . \n Using this, . Or .

The even part of any signal is given by the formula . \n Given , we first find . \n Now, we substitute these into the formula for the even part: \n . This function is non-zero for both positive and negative time.

This integral is solved using the sifting property of the Dirac delta function, which states . \n In this case, and the impulse is located at . \n We evaluate at : \n . \n Therefore, the value of the integral is 8.

Sampling is performed by replacing with . \n . \n The discrete-time frequency is . Since this frequency is outside the principal range , aliasing occurs. The observed frequency is found by subtracting . \n . \n Since , the resulting signal is .

The convolution of a rectangular pulse with itself results in a triangular pulse. The duration of the resulting signal is the sum of the durations of the individual signals, which is . The peak amplitude of the resulting triangle is , where A is the amplitude of the rectangle (here A=1).

The energy of is . \n Let's use a substitution , which means and . \n The integral becomes . \n . \n The integral part is the definition of the energy of , which is . \n Therefore, . Time compression () reduces energy, and time expansion () increases energy.

A discrete-time complex exponential is periodic if and only if its frequency is a rational multiple of . The period is the smallest integer such that for some integer . \n Here, . \n So, . \n We need to find the smallest positive integer that makes an integer. Choosing gives the smallest integer . \n .

The operation is a combination of time reversal and time shifting. Let's evaluate for a few values of : \n . \n . \n . \n . \n . \n The sequence is a time-reversed version of shifted to the right by 3 samples. The original sequence is $x[n] = {1

The unit ramp function is . We evaluate the expression at . \n . \n We evaluate each term: \n . \n . \n . \n Substituting these values back: .

The original signal is for . The transformation is . \n First, find the new interval. Let . The interval for is . \n . \n Second, find the new function. . Since , we have on this new interval. \n The function is a line with slope 2. Thus, the transformed signal is a ramp with slope 2 for .

A signal is a power signal if its average power is finite and non-zero (), which implies it has infinite energy. \n A) is an exponentially decaying signal. It has finite energy, so it's an energy signal. \n B) The sum of two sinusoids with different frequencies is periodic. Any periodic signal is a power signal. Its power is finite and its energy is infinite. \n C) is a rectangular pulse of finite duration. It has finite energy, so it's an energy signal. \n D) is a ramp signal. Its energy and power are both infinite. It is neither an energy nor a power signal.

The odd component of a signal is given by . \n Given , we have . \n So, . \n Let's evaluate this for different values of n: \n For : and . So, . \n For : and . So, . \n For : and . So, . \n This corresponds to the description in the correct option.

We use the product rule for differentiation: . \n Let and . \n Then and the generalized derivative of the step function is . \n Applying the rule: . \n Using the property , we simplify the second term: . \n Combining the terms, the result is .

A discrete-time sinusoidal signal is periodic if and only if its frequency is a rational multiple of . That is, for some integers and . \n In this case, . \n We check the ratio: . \n Since is an irrational number, this ratio is irrational. Therefore, the signal is not periodic.

The signal can be separated into its real and imaginary exponential parts using Euler's formula: \n . \n The term is a real exponential with a negative exponent, which represents exponential decay. \n The term represents a sinusoidal oscillation with frequency rad/s. \n The combination of these two terms results in a sinusoid whose amplitude, , is decaying over time.

One of the properties of convolution is that for a linear time-invariant system, . \n Let's consider the convolution with the delta function itself: . \n Now, let's use the property with . \n is the convolution of with the derivative of . This is equivalent to the derivative of the convolution of and . \n So, .

We can perform the convolution using the sliding/tabular method. The length of the resulting sequence will be . \n . \n . \n . \n . \n The resulting sequence is .

The transformation is . This is a time reversal followed by a right shift of 2, or a left shift of 2 followed by a time reversal. Let's evaluate at several points: \n . \n . \n . \n . \n The new sequence has the value 4 at , 3 at , 2 at , and 1 at . So the sequence is where the underline is at n=2. Re-evaluating. . This sequence is for . This corresponds to the option where the values for are present, but the position of the underline seems to be shifted. Let's recheck the options. The option means . This is not what we calculated. Let's re-calculate: \n . \n . \n . \n . \n . \n The sequence is where 4 is at . So, . There must be a typo in the options provided. The closest option would imply a different transformation or initial sequence. Given standard question formats, the most likely intended answer is a re-indexed sequence. If , this is reversed and shifted. The sequence of values is correct, but their index is debatable based on the options. Let's assume the question asks for the values in the resulting non-zero sequence. The sequence of values is 4, 3, 2, 1. Option A: has the correct order of values but the index is shifted. Let's choose the option with the correct reversed order of values. Re-evaluating the correct option A: . This means . But in the problem, . There is an inconsistency. The correct answer should be a sequence containing with 4 at . Let's assume a typo in the question: . Then . No. Let's assume the correct answer option A has a typo and should be which is . Let's write a new question instead of this broken one.

For a sum of two periodic signals to be periodic, the ratio of their fundamental periods (or frequencies) must be a rational number. The first term, , has a frequency Hz. The second term, , has an angular frequency rad/s, so its frequency is Hz. The ratio of the frequencies is . Since this ratio is an irrational number, the combined signal is not periodic.

The signal is non-zero when its argument is between -1 and 2. For the signal , the argument is . Therefore, we set up the inequality: To solve for , we first subtract 4 from all parts: Now, we divide by -2. Remember to reverse the inequality signs when dividing by a negative number: So, is non-zero for the interval .

The signal is a ramp from to . It is 0 elsewhere. The energy of is . The signal is a ramp from to . It is non-zero only for . The signal is the odd part of . Since is non-zero on and is non-zero on , their supports are disjoint except at . The energy of is . Because they don't overlap, this is . The energy of is the same as , which is . Therefore, the total energy is .

This problem uses the sifting property of the derivatives of the Dirac delta function. The general property is: where is the n-th derivative of evaluated at . In this case, , , and . We need to find the second derivative of : Now, we evaluate this second derivative at : Plugging this into the formula, the value of the integral is:

The derivative of an even function is an odd function, and the derivative of an odd function is an even function. We start with . Then . Let's analyze the derivative of the even part, . We check its symmetry: . So, is an odd function. Similarly, let's analyze the derivative of the odd part, . We check its symmetry: . So, is an even function. Since is the sum of an odd part and an even part, we must have and .

For a sum of discrete-time sinusoids to be periodic, each component must be periodic. The period of the overall signal is the least common multiple (LCM) of the individual periods. For the first term, , the frequency is . The period is given by , so . The smallest integer is 24 (for ). For the second term, , the frequency is . The period is given by , so . The smallest integer is 16 (for ). The fundamental period of the combined signal is the LCM of the individual periods: .

First, we find the derivative signal . Since differentiation is a linear operation and , we have: Let's analyze . For , . For , . For , . For , . So, is a signal that is +1 on and -1 on . The total energy of is given by .

The sampling period is s. The discrete-time signal is . The frequencies in the discrete-time signal are , where is the continuous-time angular frequency. For the first component, , so . For the second component, , so . Discrete-time frequencies are unique only in the range . The frequency is outside this range. We find its alias by subtracting : . So, the resulting discrete-time signal has components with frequencies and (since ). The periods are and , which gives a fundamental period of . The overall period is . The fundamental angular frequency is .

First, let's classify it as an energy or power signal. The magnitude is . This magnitude is bounded between 0 and 2, but does not decay to zero. Therefore, the signal has infinite energy and finite power, making it a power signal. Second, let's determine its periodicity. The sum of two periodic signals is periodic if and only if the ratio of their periods (or frequencies) is a rational number. Here, the ratio of the angular frequencies is given as irrational. Therefore, the signal is not periodic. However, since it is composed of periodic components, it is termed quasi-periodic. Thus, it is a quasi-periodic power signal.

The vector step will be a series of 0s followed by a transition to a series of 1s. The transition happens at t=0. Let's look at the values around this point: step will look like [..., 0, 0, 1, 1, ...]. The np.diff(step) operation computes the difference between adjacent elements, so it will be [..., 0, 1, 0, ...]. There will be a single value of 1 at the point of transition, and 0 everywhere else. The simulated impulse is then calculated as np.diff(step) / dt, where dt=0.01. The resulting array will be mostly zeros, with one element having the value 1 / 0.01 = 100. This demonstrates how a continuous-time impulse, with infinite height and zero width, is approximated in a discrete simulation as a pulse with a large but finite height and a width equal to the time step.

First, we find the expression for the integrand, which is the transformed input signal. The input is , so we need to evaluate . We use the scaling property of the Dirac delta function: . Applying this, we get: Now, we substitute this back into the system equation: The integral of the delta function is the unit step function . Therefore, the output is:

The signal can be expressed in terms of the even and odd parts of . We know that . Also, . By definition, and . Substituting these into the expression for gives . Now we can find : The energy of is . Given that the energy of the even part , the energy of is . The energy of the odd part, , is extraneous information.

A discrete-time complex exponential signal is periodic if and only if its angular frequency is a rational multiple of . This can be expressed as the condition that must be a rational number. In this case, . We check the ratio: Since is an irrational number, is also irrational. Therefore, the condition for periodicity is not met, and the signal is not periodic. This is a key difference from continuous-time signals , which are always periodic for any non-zero .

The average power of a signal is . For a complex signal, . Here, , so its complex conjugate is . The average power of is therefore: The first term is by definition . The second term is also , because a time shift does not change the average power of a signal. Thus, .

The convolution integral is . We need to evaluate this at : . The signal is a rectangular pulse, equal to 1 for and 0 otherwise. This limits the integration bounds: . Now we analyze the signal . For . For . For . So is a ramp up to 1, and then stays at 1. The argument of in the integral is . We need to evaluate this for in the range . When , the argument is 4. When , the argument is 2. For the entire integration range , the argument is always greater than or equal to 2. In this region, . So the integral becomes: .

The signal is a train of rectangular pulses of width 1. The term represents a pulse of width 1 centered at . The term means the pulses alternate in sign. For , we have a positive pulse centered at . For , a negative pulse centered at . For , a positive pulse centered at , and so on. The signal pattern repeats every two pulses, for a change in time of $4$. Thus, the fundamental period is . To find the average power, we calculate the energy over one period and divide by the period length: . Let's choose the period from to . In this interval, we have a full positive pulse from to (from ) and a full negative pulse from to (from ). The magnitude squared will be during the first pulse and during the second pulse. The total duration within one period where is . The energy in one period is . The average power is .

The signal is a decimated version of . A signal is periodic if is periodic with period N. The new period is given by the formula , where gcd is the greatest common divisor. Here, and the decimation factor is . The greatest common divisor of 3 and 15 is . Therefore, the fundamental period of is . The intuitive reason is that as you step through for , you are picking samples from the original signal. The pattern in repeats every 15 samples. The new sequence will repeat when is a multiple of 15, i.e., . The smallest integer for this is (when ).

Let's define a new signal . We need to determine the symmetry of . By definition, an even signal has the property , and an odd signal has the property . Let's evaluate : This shows that is an odd signal. The summation of any odd signal over all symmetric limits from to is always zero (assuming the sum converges). This is because for every term in the sum, there is a corresponding term , and they cancel each other out. Also, for an odd signal, must be 0. Thus, the summation is zero regardless of the specific signal .

The signum function, , is defined as 1 for , -1 for , and 0 for . It can be expressed in terms of the unit step function as . Now, we differentiate this expression. The derivative of a constant is zero, and the derivative of the unit step function is the unit impulse function, . First derivative: Second derivative: The result is twice the unit doublet (the derivative of the unit impulse).

The energy of a signal is given by . The energy of is: To solve the integral, we use a change of variables. Let . Then and . We must consider the sign of B. To handle both positive and negative B, we use . If , the limits remain . If , the limits are swapped but the negative sign from swaps them back. So we can write: The integral is the definition of the energy of , which is . Therefore: Time compression () concentrates the signal, increasing its energy. Time expansion () spreads it out, decreasing its energy.