Question

Define a rectangular pulse p(t) = u(t+1)-u(t-1) = { 1, -1≤ t ≤1 0, otherwise 1. An LTI system has an impulse response h(t) = tu(t-5). If the input is x(t) = t²[u(t-1) - u(t-3)], find the output. 2. Determine whether the contintuous-time LTI systems characterized by the following impulse responses are causal or non-causal, stable or non- stable. (a) h(t) = e^tu(-t) (b) h(t) = (-t)e-tu(-t) (c) h(t) = e-12t (d) h(t) = p(t/2). (e) h(t) = d(t) + e-stu(t) 3. Are the LTI systems with the following impulse responses invertible? If invertible, find the inverse system. (a) h(t) = 3d(t+3) (b) h(t)=(t-3) + (t - 5). 4. Consider a circuit with a voltage source, v(t), a resistor with resistance R, and a capacitor with capacitance C connected in serise. If the input of the system is the voltage source v(t), and the output of the system is the voltage across the capacitor, v(t). Write the system equation in the form of a differential equation.

          Define a rectangular pulse p(t) = u(t+1)-u(t-1) = 
{
1, -1≤ t ≤1
0, otherwise
1. An LTI system has an impulse response h(t) = tu(t-5). If the input
is x(t) = t²[u(t-1) - u(t-3)], find the output.
2. Determine whether the contintuous-time LTI systems characterized by
the following impulse responses are causal or non-causal, stable or non-
stable.
(a) h(t) = e^tu(-t)
(b) h(t) = (-t)e-tu(-t)
(c) h(t) = e-12t
(d) h(t) = p(t/2).
(e) h(t) = d(t) + e-stu(t)
3. Are the LTI systems with the following impulse responses invertible?
If invertible, find the inverse system.
(a) h(t) = 3d(t+3)
(b) h(t)=(t-3) + (t - 5).
4. Consider a circuit with a voltage source, v(t), a resistor with resistance
R, and a capacitor with capacitance C connected in serise. If the input
of the system is the voltage source v(t), and the output of the system
is the voltage across the capacitor, v(t). Write the system equation in
the form of a differential equation.

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define a rectangular pulse pt ut1 ut 1 1 1 t 1 0 otherwise 1 an lti system has an impulse response ht tut 5 if the input is xt t2ut 1 ut 3 find the output 2 determine whether the contintuous 85258

Added by Clara G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Adi S

07/30/2024

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Step 1: The output of an LTI system is the convolution of the input and the impulse response. Show more…

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Define a rectangular pulse p(t) = u(t+1)-u(t-1) = { 1, -1≤ t ≤1 0, otherwise 1. An LTI system has an impulse response h(t) = tu(t-5). If the input is x(t) = t²[u(t-1) - u(t-3)], find the output. 2. Determine whether the contintuous-time LTI systems characterized by the following impulse responses are causal or non-causal, stable or non- stable. (a) h(t) = e^tu(-t) (b) h(t) = (-t)e-tu(-t) (c) h(t) = e-12t (d) h(t) = p(t/2). (e) h(t) = d(t) + e-stu(t) 3. Are the LTI systems with the following impulse responses invertible? If invertible, find the inverse system. (a) h(t) = 3d(t+3) (b) h(t)=(t-3) + (t - 5). 4. Consider a circuit with a voltage source, v(t), a resistor with resistance R, and a capacitor with capacitance C connected in serise. If the input of the system is the voltage source v(t), and the output of the system is the voltage across the capacitor, v(t). Write the system equation in the form of a differential equation.

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00:01 We are given that there is an lti system.

00:03 For this lti system, the input signal that is given to us is input signal is that xn is equal to u .n.

00:15 Minus 3.

00:17 Similarly the impulse response impulse response is given that hn is equals to 0 .8 raised to the power n u n minus 2.

00:38 Now to find this we need to plot the value so we know that y n is equal to the summation of k is equals to minus infinity up till infinity u n minus 3 0 .8 raised to the power n minus k u n minus k minus 2 this becomes your equation now for this equation we know that n minus 2 is either greater than or equal to 3 so we can modify this according to this condition therefore y n will be equal to summation from k is equal to 3 n minus 2 0 .8 raised to the power n minus k that is equals to 0 .8 raised to the power n summation of k is equal to 8 to n minus 2 0 .8 raised to the power minus k that is equals to 0 .8 raised to the power n k is equals to 8 up till n minus 2 1 .25 raised to the power k.

01:46 This becomes your y n.

01:48 Mark this as our equation number 1.

01:51 Now similarly, we can write this in a generalized form.

01:57 Generalized form, that is summation from k is equals to n1 up till n2, a raised to the power n is equals to a raised to the power n minus a n2 plus 1 divided by 1 minus 8.

02:12 Now this is your generalized equation.

02:15 So in this equation, we can now put the value.

02:18 And solve for output.

02:21 Therefore, yn will be equal to 0 .8 raised to the power n into 1 .25 whole cube minus 1 .25 n minus 2 plus 1.

02:33 That is the power, divided by 1 minus 1 .25.

02:37 That is equals to 0 .8 raised to the power n then 0 .8 raised to the power minus 3 minus 0 .8 into 0 .8 raised to the power minus n divided by minus 0 .25...

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