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Please fill out the table. Do not copy from other answers already on Chegg, thank you. Prelab Assignment 3: Model each signal x(t) by an even function of time and find the exponential form of its Fourier coefficients Xn, which should be real and even functions of n. Compute the first 9 exponential Fourier coefficients Xn, 1 < n < 9, of each of the periodic waveforms in Table 3 and enter the results in that table. Start with the expressions for H(t) given for the single pulse versions in Table 2, and from these derive the Exponential Fourier Series coefficients denoted as Xn for the periodic versions, using the relationships between Fourier Series and Fourier Transforms shown above. Use the specific period (T) and duration specified in Table 3 when computing the Fourier Series coefficients for your table. Note that for Table 3: i. duration refers to the on-time (i.e., the duration of the pulse) in Table I ii. The fundamental frequency fo = 1/T iii. The angle expression n * fo * t has units of "radians", not "degrees". Table 3a. Exponential Fourier coefficients of the first 9 harmonics of periodic waveforms x( X1 X2 X3 X4 X5 X6 X7 X8 X9 Periodic square Vbase-to-peak = 1V Duration = 500 sec Period = 1 msec Periodic rectangle Vbase-to-peak = 1V Duration = 200 sec Period = 1 msec Periodic triangle Vbase-to-peak = 1V Duration = 1 msec Period = 1 msec

          Please fill out the table. Do not copy from other answers already on Chegg, thank you.

Prelab Assignment 3: Model each signal x(t) by an even function of time and find the exponential form of its Fourier coefficients Xn, which should be real and even functions of n. Compute the first 9 exponential Fourier coefficients Xn, 1 < n < 9, of each of the periodic waveforms in Table 3 and enter the results in that table.

Start with the expressions for H(t) given for the single pulse versions in Table 2, and from these derive the Exponential Fourier Series coefficients denoted as Xn for the periodic versions, using the relationships between Fourier Series and Fourier Transforms shown above. Use the specific period (T) and duration specified in Table 3 when computing the Fourier Series coefficients for your table. Note that for Table 3: i. duration refers to the on-time (i.e., the duration of the pulse) in Table I ii. The fundamental frequency fo = 1/T iii. The angle expression n * fo * t has units of "radians", not "degrees".

Table 3a. Exponential Fourier coefficients of the first 9 harmonics of periodic waveforms
x(
X1
X2
X3
X4
X5
X6
X7
X8
X9
Periodic square Vbase-to-peak = 1V Duration = 500 sec Period = 1 msec
Periodic rectangle Vbase-to-peak = 1V Duration = 200 sec Period = 1 msec
Periodic triangle Vbase-to-peak = 1V Duration = 1 msec Period = 1 msec
        
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please fill out the table do not copy from other answers already on chegg thank you prelab assignment 3 model each signal xt by an even function of time and find the exponential form of its  92898

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Please fill out the table. Do not copy from other answers already on Chegg, thank you. Prelab Assignment 3: Model each signal x(t) by an even function of time and find the exponential form of its Fourier coefficients Xn, which should be real and even functions of n. Compute the first 9 exponential Fourier coefficients Xn, 1 < n < 9, of each of the periodic waveforms in Table 3 and enter the results in that table. Start with the expressions for H(t) given for the single pulse versions in Table 2, and from these derive the Exponential Fourier Series coefficients denoted as Xn for the periodic versions, using the relationships between Fourier Series and Fourier Transforms shown above. Use the specific period (T) and duration specified in Table 3 when computing the Fourier Series coefficients for your table. Note that for Table 3: i. duration refers to the on-time (i.e., the duration of the pulse) in Table I ii. The fundamental frequency fo = 1/T iii. The angle expression n * fo * t has units of "radians", not "degrees". Table 3a. Exponential Fourier coefficients of the first 9 harmonics of periodic waveforms x( X1 X2 X3 X4 X5 X6 X7 X8 X9 Periodic square Vbase-to-peak = 1V Duration = 500 sec Period = 1 msec Periodic rectangle Vbase-to-peak = 1V Duration = 200 sec Period = 1 msec Periodic triangle Vbase-to-peak = 1V Duration = 1 msec Period = 1 msec
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Transcript

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00:01 So, in this question we have given with two signals that is xt is given us like this.
00:09 We can draw it.
00:10 So, the responses are here.
00:12 This is minus pi.
00:13 This is pi.
00:15 Again, it is getting repeated at 10 pi or the midpoint is 10 pi.
00:24 Here it is minus 10 pi.
00:27 So, we can find the period of this signal.
00:33 This is your xt.
00:34 So, period of this signal we can find out that will be t equal to 10 pi.
00:44 So, omega naught will be 2 pi upon t.
00:48 That will be 2 pi by 10 pi which comes out to be 1 by pi.
00:58 So, xt we can express in terms of its fourier coefficient that is n tending to minus infinity to infinity summation of cn e to the power j omega naught nt.
01:11 So, in that case cn can be or the fourier coefficient, fourier series coefficient can be calculated as xt e power minus j omega nt dt.
01:30 So, putting these values here we can say that since it is a square signal we can say that this will be xt will be 1 and its amplitude is 1 and we have our time period that is minus pi to pi and here it is 1.
01:51 So, e to the power minus j omega naught nt dt.
01:57 So, we know that integration of e power x is e power x.
02:03 So, here it will come out as e to the power minus j omega naught nt upon minus j omega naught n.
02:17 That is the coefficient of t in the expression.
02:21 So, here it is limits are minus pi to pi.
02:24 So, further solving it we can get the values like this.
02:28 So, we can get the values but here time period we can substitute is 1 by 10 pi and in the inside here values will come out to be e to the power minus j omega naught n pi minus e to the power minus j omega naught n minus pi.
02:51 So, this will become plus pi and in the denominator we have minus j omega naught n.
03:00 So, putting the values of omega naught and rewriting them we can get the rearranged value like this e to the power j n j e to the power j n pi by pi and here it will be minus e to the power minus j n pi by pi whole divided by this minus term actually get rearranged in the numerator.
03:35 So, we will get n pi by pi n by pi.
03:42 So, this pi get cancelled with 10 and here we get 2.
03:48 So, we can rewrite our expression as let's see what we can write.
03:57 So, we can rewrite it e to the power j n pi by pi minus e to the power minus j n pi by pi whole divided by j 2 we can write and outside we will be having 1 by n pi or 1 by pi n we can write.
04:26 So, this term is the general expansion of sin theta.
04:31 So, we can write directly that this is that your c n value will be that's implies c n or fourier series coefficient for x t will be 1 by pi n sin n pi by pi.
04:49 So, this is the fourier series coefficient of x t.
04:53 So, we can finally write the fourier series expansion as c 0 c 1 c 0 c 1 c 2 etcetera can be calculated with n equal to 0 1 2 putting the values we can calculate...
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