# How Do You Analyze a Continuous Time LTI System Using Differential Equations?

• crom1
In summary: I think that's a sin.In summary, we have a continuous time LTI system with a differential equation y'(t) + 5y(t) = 10u(t). We found the transfer function to be Y(s) = 10/(s+5), with a pole at s=-5 and Re(s)<0, indicating that the system is stable. By plugging in s=jω, we determined the frequency characteristics to be |H(jω)| = 10/(√s^2+ω^2) and ∠H(jω) = -arctan(ω/5). Then, using Laplace transformation, we found the impulse response to be y(t) = 10*e^(-
crom1

## Homework Statement

continuous time LTI system is given with differential equation y'(t) + 5y(t) = 10u(t)
a) Find transfer function of system and determine is it stable or not.
b)Find frequency characteristics (amplitude and phase angle) of given system
c) Find impulse response using Laplace transformation
d) Find zero-input response(I think that's the right word) for u(t) = 6cos(5t)μ(t)

## The Attempt at a Solution

Ok, I got
a) Y(s) = 10/(s+5) * U(s) , pole is s=-5 , and Re(s)<0 so the system is stable
b) Plug s=jω in H(s)=10/(s+5) , then |H(jω)| = 10/(√s^2+ω^2) and ∠H(jω) = - arctan(ω/5)
c) for impulse response U(s) = 1, Y(s) = 10/(s+5) ⇒ y(t) = 10*e^(-5t) , t>0
d) Using frequency characteristics, we have ω=5, and |H(j5)| = √2 and ∠H(j5) = -π/4 so the response is
y(t) = 6√2 *cos(5t-pi/4)

I need someone to say if these are correct or not, if not i will post my full attempt and reasoning. Thanks

crom1 said:
Ok, I got
a) Y(s) = 10/(s+5) * U(s) , pole is s=-5 , and Re(s)<0 so the system is stable
10 U(s) is not part of the transfer function. It's a specific input; the transfer function gives output for any input.
b) Plug s=jω in H(s)=10/(s+5) , then |H(jω)| = 10/(√s^2+ω^2) and ∠H(jω) = - arctan(ω/5)
OK but leave out the "10". Also, typo used "s" for "5".
c) for impulse response U(s) = 1, Y(s) = 10/(s+5) ⇒ y(t) = 10*e^(-5t) , t>0
U is a step function.The symbol for impulse input is δ(t). Your actual input is kδ(t), k = 1 volt-sec. Careful with δ(t), its units are time-1. Again, ditch the "10".
d) Using frequency characteristics, we have ω=5, and |H(j5)| = √2 and ∠H(j5) = -π/4 so the response is
y(t) = 6√2 *cos(5t-pi/4)
never heard of "zero input response". You mean mean input is 6cos(5t)U(t) or ? Never saw μ(t) either.
Strange they never asked for the time response to the 10u(t) input ...

## 1. What is a signal in a signals and systems problem?

A signal in a signals and systems problem is a variable that carries information in the form of a physical quantity. It can be represented graphically as a function of time or space.

## 2. What is a system in a signals and systems problem?

A system in a signals and systems problem is a mathematical model that processes the input signal to produce an output signal. It can be represented by a block diagram or a set of equations.

## 3. What is the difference between continuous-time and discrete-time signals and systems?

Continuous-time signals and systems operate on signals that are defined over a continuous domain, such as time. Discrete-time signals and systems operate on signals that are defined at discrete time instances, such as samples taken at regular intervals.

## 4. How do I analyze a signals and systems problem?

To analyze a signals and systems problem, you can use mathematical tools such as Fourier transforms, Laplace transforms, and Z-transforms. These tools allow you to understand the behavior of a system and its response to different types of signals.

## 5. What are some real-world applications of signals and systems?

Signals and systems have a wide range of applications, including telecommunications, image and signal processing, control systems, and biomedical engineering. They are used to analyze and process signals in various fields to extract information and make decisions.

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