Homework 1 Solutions

Transcription

1 18-9 Signals and Systems Profs. Byron Yu and Pulkit Grover Fall 18 Homework 1 Solutions Part One 1. (8 points) Consider the DT signal given by the algorithm: x[] = 1 x[1] = x[n] = x[n 1] x[n ] (a) Plot the signal x[n] for n 11. (b) Determine if the signal is periodic, and if so find its fundamental period. Solution: (a) Plug different values of n into the given equation to calculate x[n] for n 11 x[] = x[1] x[] = 1 = 1 x[3] = x[] x[1] = 1 = 1 x[4] = x[3] x[] = 1 1 = x[5] = x[4] x[3] = ( 1) = 1 x[6] = x[5] x[4] = 1 ( ) = 1... Using the values we get we can draw the plot below. 1 x[n] n (b) From the plot above we know that x[n] is periodic with a fundamental period of 6.

2 Homework 1 Solutions Below is a proof if you are interested. For any n Z x[n + 6] = x[n + 5] x[n + 4] = x[n + 4] x[n + 3] x[n + 3] + x[n + ] = x[n + 4] x[n + 3] + x[n + ] = x[n + 3] x[n + ] x[n + 3] + x[n + ] = x[n + 3] = x[n + ] + x[n + 1] = x[n + 1] + x[n] + x[n + 1] = x[n] Therefore the signal x[n] is periodic with a period of 6. Note that 6 = 3 but from the plot we know x[n + ] x[n] and x[n + 3] x[n] for at least some values of n. Since x[] x[1] (meaning that the fundamental period is not 1), the fundamental period of the signal x[n] is 6.

3 Homework 1 Solutions 3. (8 points) Consider the CT signal given by the fomula: x(t) = cos (πt/3) (a) Plot the signal x(t) for 3 t 3. (b) Determine if the signal is periodic, and if so find its fundamental period. Solution: (a) Recall that cos(θ) = cos (θ) 1 Therefore for x(t) we have x(t) = cos (πt/3) = cos(4πt/3) + 1 So the plot of x(t) is a cosine wave shifted up by 1. x(t) t (b) From the plot we know that the signal x(t) is periodic with a fundamental period of 1.5. Below is a proof if you are interested. Assume x(t) is periodic with period T. Then t R, x(t) = x(t + T ). cos(4πt/3) + 1 = cos(4π/3(t + T )) + 1 cos(4πt/3) = cos(4πt/3 + 4πT/3) 4πT/3 = kπ(k Z) T = 3k/(k Z) Therefore x(t) is periodic, and the fundamental period is 1.5.

4 4 Homework 1 Solutions 3. (1 points) Determine if each of the following signals is periodic and, if so, find its fundamental period: (a) x(t) = sin (πt) (b) x(t) = e t cos(πt) (c) x(t) = cos( π 3 t) + sin( π t) (d) x(t) = cos(πt) + cos( πt) Solution: (a) x(t) = sin (πt) The signal is periodic with a fundamental period of 1. Observe that, sin (π(t + 1 )) = sin (πt + π) = ( sin(πt)) = sin (πt) This shows that 1 is a period but it is not enough to show that it is the fundamental period, i.e. the smallest c > such that x(t + c) = x(t) for all t. But observe that, sin (πt) = 1 cos(4πt) The fundamental period of cos(t) is π. Thus, the fundamental period of cos(4πt) is π = 1. 4π (b) x(t) = e t cos(πt) The signal is not periodic. This holds since e t is a monotonically decreasing function. Suppose there exists a c > such that x(t + c) = x(t) for all t. Then we must have e t cos(πt) = e (t+c) cos(π(t + c)) However, this implies that e c = cos(π(t+c)) for all t which is not possible since c is cos(πt) a constant while cos(π(t+c)) can take different values for different values of t. cos(πt) (c) x(t) = cos( πt) + sin( πt) 3 The signal is periodic with a fundamental period of 1. As this signal has two parts, we need to see if both of these parts are periodic firstly. For cos( πt), 3 it is periodic with a fundamental period of 3. For sin( π t), it is periodic with a fundamental period of 4. Next, we need to find out if there exist least common multiple from 3 and 4. It is 1! We find out the fundamental period of x(t) equals 1! (d) This signal is not periodic. Although the two parts have fundamental period of 1 and, we cannot find least common multiple for 1 and.

5 Homework 1 Solutions 5 4. (1 points) (a) Consider the following DT signal: x[n] = {( 1 ) n n otherwise Determine the total energy and average power of x[n]. (b) Consider the following CT signal: 3t t 5 x(t) = 15 5 < t 1 otherwise Determine the total energy and average power of x(t). Solution: (a) Total energy of x[n] is To find the total energy, E = + n= x[n] = x[n] since Signal is elsewhere. n= ( ) n 1 = = n= = 1 ( 1 4 ) n= = ( ) n 1 4 Average power of x[n] is. To find the average power, 1 P = lim N N + 1 N n= N x[n] = lim N (b) Total energy of x(t) is 15. To find the total energy, E = + x(t) dt = = 1 5 [ t 3 = (N + 1) (3 4 ) = x(t) dt since Signal is elsewhere. (3t) dt + ] (15 )[t] 1 5 = = 15 (15) dt

6 6 Homework 1 Solutions Average power of x(t) is. To find the average power, 1 P = lim T T +T T x(t) 1 dt = lim T T 15 = Explanation: Suppose you choose any T > 1, i.e. beyond the range where x(t) is non-zero. Observe that, Thus, E = = + +T T x(t) dt = T x(t) dt }{{} = as signal is x(t) dt for any arbitrary T > 1 1 P = lim T T +T T +T + x(t) dt + T x(t) 1 dt = lim T T E = x(t) dt +T }{{} =as signal is

7 Homework 1 Solutions 7 5. (1 points) Decompose the following signals into their odd and even components. Simplify your answer to the simplest expression: (a) x(t) = 1t 5 + 4t 4 t 3 t 1 (b) x(t) = 1 + t cos(t) + t 3 sin (t) + t 4 cos(t) sin(t) (c) x[n] = cos(n) + sin(n) + ( 1) n + ( 1) n sin( π 7 n) (d) x(t) = sin(3t) sin(t) sin(t) (Hint: sin( x) = sin(x)) Solution: (a) x(t) = 1t 5 + 4t 4 t 3 t 1 Odd: 1 (x(t) x( t)) = 1t5 t 3 Even: 1 (x(t) + x( t)) = 4t4 t 1 Observe that for a general polynomial, the odd powers belong to the odd part and the even powers belong to the even part of the signal. (b) x(t) = 1 + t cos(t) + t 3 sin (t) + t 4 cos(t) sin(t) Odd: 1 (x(t) x( t)) = t cos(t) + t3 sin (t) + t 4 cos(t) sin(t) Even: 1 (x(t) + x( t)) = 1 (c) x[n] = cos(n) + sin(n) + ( 1) n + ( 1) n sin( π 7 n) Odd: 1 (x(t) x( t)) = sin(n) + ( 1)n sin( π 7 n) Even: 1 (x(t) + x( t)) = cos(n) + ( 1)n (d) x(t) = sin(3t) sin(t) sin(t)

8 8 Homework 1 Solutions Odd: 1 (x(t) x( t)) = 1 (sin(3t) sin(t) sin(t) sin( 3t) sin( t) sin( t)) = 1 (sin(3t) sin(t) sin(t) ( 1)3 sin(3t) sin(t) sin(t)) = sin(3t) sin(t) sin(t) Even: 1 (x(t) + x( t)) = 1 (sin(3t) sin(t) sin(t) + sin( 3t) sin( t) sin( t)) = 1 (sin(3t) sin(t) sin(t) + ( 1)3 sin(3t) sin(t) sin(t)) =

9 Homework 1 Solutions 9 Part Two 6. (1 points) A pulse x(t) is defined by: x(t) = (a) Determine the total energy of x(t). { A cos(ωt) sin(ωt) t T otherwise (b) A differentiator is applied to x(t), defined by: y(t) = d dt x(t) Determine the resulting output y(t) of the differentiator. energy of y(t). Determine the total Solution: (a) [5 points].determine the total energy of x(t) Since we have A cos(ωt) sin(ωt) = 1 A sin(ωt), E = T = (A cos(ωt) sin(ωt)) dt T = A 4 ( 1 A sin(ωt)) dt T Using (sin(ωt)) = 1 (1 cos(4ωt)) we have, (sin(ωt)) dt E = A 4 = A 8 T T 1 cos(4ωt) dt 1 cos(4ωt)dt T = A 8 T A cos(4ωt)dt 8 = A 8 T A 1 8 4ω sin(4ωt) T = A 8 A 3ω sin(4ωt )

10 1 Homework 1 Solutions (b) [5 points].determine the resulting output y(t) of the differentiator. Determine the total energy of y(t). Differentiating x(t) respect to t, For the total energy of y(t), E = T y(t) = aω sin (ωt) + Aω cos (ωt) = Aω(cos (ωt) sin (ωt)) { Aω cos(ωt) t T = otherwise (Aω cos(ωt)) dt = = T = T 1 a ω dt + T A ω cos (ωt)dt A ω ( 1 + cos(4ωt) )dt T 1 A ω cos(4ωt)dt = 1 A ω T + A ω 1 8ω sin(4ωt) T = 1 A ω T + A ω 8 sin(4ωt )

11 Homework 1 Solutions (1 points) MATLAB Commands Please download MATLAB from the following website, which also provides instructions on how to do so. ECE students can choose to install the standalone edition in order to use MATLAB off campus. When you start MATLAB, there are two windows to note: a workspace, which contains the current variables; a command window, which allows for the input of commands. When MATLAB is first started, both should be empty except for the command window prompt. >> You can create variables like in other programming languages: >> a = You will note that the variable a now appears in the workspace. Once you have created variables, they will be saved in the workspace and you can use them in your new commands without having to define them again. Square brackets are used to create matrices. Columns are separated by spaces (or commas), and rows are separated by semicolons. Try the following commands in your MATLAB command window. To get information about any command, you can type help <name of the command>. >> b = [1 3 5] >> c = [1; 3; 4] >> d = [1 3; 5 6 7] Try the following MATLAB operators: >> b*c >> b+b >> b+c >> d*c Note that if the dimensions do not agree, you cannot use certain operators. MATLAB has a large library of commands that help to manipulate your matrices. Try the following commands, and briefly explain what each of the commands do. Note that MATLAB indexing starts at 1, rather than in many other programming languages. >> a = zeros (3, 4) >> b = ones (, 5)

12 1 Homework 1 Solutions >> c = eye (4) >> d = size (a) >> abs ([ -5, 3]) >> ceil (3.4) >> help floor >> e = [:3:1] >> f = e >> g = e () >> h = cos (pi /) >> k = exp (1.) >> ex1 = linspace (1, 5, 4) >> ex1 = [ ex1 ; ex1 *] >> n = ex1 (1, ) >> p = ex1 (1, :) >> m = ex1 (1, : end ) >> q = [p m] >> cmplx = 3 - i >> real ( cmplx ) >> imag ( cmplx ) >> abs ( cmplx ) >> angle ( cmplx ) >> 5^ >> 3 == 3 >> 3 == >> 3 ~= 1 >> whos >> clear a b There are many other built-in functions in MATLAB. MATLAB has excellent documentation, so there are plenty of online resources for helping you find the functions you need. Solution: (1) create a 3x4 matrix of zeros () create a x5 matrix of ones (3) create a 4x4 identity matrix (4) get the dimensions of matrix a (5) take the absolute values of each element in the matrix [-5, 3] (6) round to the smallest integer that is greater than 3.4 (7) retrieve documentation on using floor (8) create a vector of values starting from and counting by 3 until 1

13 Homework 1 Solutions 13 (9) take the transpose of e (1) select the second element in vector e (11) compute the cosine of π (1) compute e to the first power (13) create a linearly spaced vector of 4 values from 1 to 5 inclusive (14) multiply ex1 by and combine it with the original vector as different rows (15) select the element of ex1 in the first row, second column (16) select the first row of ex1 (17) select the first row of ex1 from the second column to the last (18) concatenate two vectors (19) create a complex value () take the real part of cmplx (1) take the imaginary part of cmplx () take the magnitude of cmplx (3) take the angle of cmplx (4) compute 5 to the second power (5) check equality of 3 versus 3 (6) check equality of 3 versus (7) check inequality of 3 versus 1 (8) display all current variables in the workspace (9) clear the variables a and b

14 14 Homework 1 Solutions 8. (1 points) MATLAB Plots Now we will get to one of the powerful tools of MATLAB: plots. In order to use the plot function, you need a vector of values for the x-axis and one for the y-axis. You can use plot(x, y) to create a continuous line plot. >> x = [ -:] >> y = [ ] >> plot (x, y) You can instead use stem(x, y) to plot discrete points in a stem plot. >> stem (x, y) Now, use these tools to make the following plots. Make sure to title your plots and label the axes. In order to do so, try help xlabel, help ylabel, and help title for instructions. (a) Odd component of y as both a continuous line plot and a discrete stem plot. (b) Even component of y as both a continuous line plot and a discrete stem plot. Solution: (a) >> x = [ : ] ; >> y = [ ] ; >> y odd =. 5 y.5 y ( 5 : 1 : 1 ) ; >> y even =. 5 y +. 5 y ( 5 : 1 : 1 ) ; >> f i g u r e >> stem ( x, y odd ) >> x l a b e l ( x ) >> y l a b e l ( y odd ) >> t i t l e ( odd component o f y ) >> f i g u r e >> p l o t ( x, y odd ) >> x l a b e l ( x ) >> y l a b e l ( y odd ) >> t i t l e ( odd component o f y ) >> f i g u r e >> stem ( x, y even ) >> x l a b e l ( x ) >> y l a b e l ( y even ) >> t i t l e ( even component o f y ) >> f i g u r e >> p l o t ( x, y even ) >> x l a b e l ( x )

15 Homework 1 Solutions 15 >> y l a b e l ( y even ) >> t i t l e ( even component o f y )

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18 18 Homework 1 Solutions 9. (1 points) Matlab Functions A square wave, which is also called a pulse wave, is a periodic function that alternates between two values. The duty cycle is the percent of the period in which the signal is at its maximum value. The figure below shows a 1 Hz square wave with equal duration at the maximum value +1 and minimum value. So, its duty cycle is 5%. Now use the functions you examined in question 8 and write your own square wave generator. The specifics are: The output variable mywave should be a square wave that has a fixed frequency of 1 Hz and alternates between 1 and. The signal lasts for 3 seconds. The input variable dutycycle ranges from to 1 and it controls the duration of the maximum value, i.e. the fraction of time for which the signal is at its maximum value, expressed as a percentage. To define a function in Matlab, click on the yellow plus sign on the top left corner of the Matlab interface and type in the code below. function mywave = WaveGenerator ( dutycycle ) % your code here end Then save and name the.m file the same name as the function name under your current working directory. That is, you should save your script as WaveGenerator.m. In your submission, provide the code for this function. A hint for this problem is to use zeros and ones command. First, lets look at the first.5 seconds with a.1 second interval. That is, lets look at the value of the square wave plotted above at time points.1,.,...,.5. The amplitude at these time points are always 1. So, to get the first half period of a square wave, use the command as follows:

19 Homework 1 Solutions 19 posvalue = ones (1,5); posvalue is a 1 5 vector that represents the first.5 seconds of the wave. Similarly, we can use zeros to generate the other half period of the wave. Then, concatenate two parts together to get a 1 1 vector for the whole period. Last, repmat function (for more information, use help repmat) allows you to repeat one period multiple times. In order to generate signals with different duty cycles, you only need to change the number of 1s and s.

20 Homework 1 Solutions Solution: function mywave = WaveGenerator ( dutycycle ) posvalue = ones (1, dutycycle ); negvalue = zeros (1,1 - dutycycle ); mywave = repmat ([ posvalue, negvalue ],1, 3); % three times of the signal end

21 Homework 1 Solutions 1 1. (9 points) Put it all together Now, it is time to use your wavegenerator and the functions you learned from question 9 and plot a pulse wave function with 5% duty cycle and 85% duty cycle. To give you an example, the code below plots the % duty cycle square wave. t =.1:.1:3; y = WaveGenerator (); plot (t,y) title ( % duty cycle square wave ) xlabel ( Time (s) ) ylabel ( Amplitude ) In your submission, you only need to show the two plots for 5% and 85% duty cycle. If you can generate plots that look like the figure above, you have already finished your first 189 homework! Yeah! But if you are a perfectionist, you can try command axis or adjust LineWidth for plot to make the plot prettier.

22 Homework 1 Solutions Solution: Figure 1 Figure