Limit of u(t)- u(t -\delta) as delta goes to zero, LTI systems

In summary, the conversation discusses finding the limit of yb(t) as \delta goes to zero. The attempt at a solution involves using L'hopital's rule and leads to the expression exp[-t/RC]/RC. However, there is uncertainty about what to do with the term u(t)-u(t-delta), and the idea of using the definition of the derivative u'(t) is suggested.
  • #1
AMac33
2
0

Homework Statement



what is the limit of yb(t) as \delta goes to zero.

yb(t) = (b/delta)*exp[-t/RC](exp[delta/RC] - 1)(u(t)-u(t-delta))

b=1.






2. The attempt at a solution

I used L'hopitals rule to find the limit of (b/delta)*exp[-t/RC](exp[delta/RC] - 1), which i got to be exp[-t/RC]/RC.

But I do not know what to do with the u(t)-u(t-delta). does it go to zero? or am I supposed to say it goes to delta(t)?

Any help/ advice would be appreciated, thanks!
 
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  • #2
Does the expression

[tex]\lim_{\delta \to 0} \frac{u(t + \delta) - u(t)}{\delta}[/tex]
ring a bell?
 
  • #3
honestly, no. I'm fairly new to the unit step and dirac delta function.
 
  • #4
Oh, but this doesn't have to do with either of those. [itex]\delta[/itex] is just a variable, if it makes you feel any better you can call it x or (more commonly used) h

Maybe I should ask it the other way around: what is the definition of the derivative u'(t) of u(t) at t?
 
  • #5




I would approach this problem by first understanding the terms and variables involved. In this case, we are dealing with a limit and an expression involving u(t) and u(t-delta) as delta goes to zero. From the given expression, we can see that b is a constant and RC is a parameter that affects the function. We also know that u(t) is the unit step function, which is equal to 1 for t >= 0 and 0 for t < 0.

Now, let's focus on the expression (u(t) - u(t-delta)). As delta goes to zero, this expression becomes 1 for all t, since u(t) is equal to 1 for all t >= 0. Therefore, we can rewrite the original expression as (b/delta)*exp[-t/RC](exp[delta/RC] - 1).

Using L'Hopital's rule, we can find the limit of this expression as delta goes to zero. This gives us exp[-t/RC]/RC. However, we cannot ignore the unit step function in the original expression. As we have established, as delta goes to zero, (u(t) - u(t-delta)) becomes 1 for all t. Therefore, the final limit of yb(t) would be 1/RC * exp[-t/RC] as delta goes to zero.

In conclusion, the limit of yb(t) as delta goes to zero is 1/RC * exp[-t/RC]. This result is valid for all values of b and RC, as long as they are not equal to zero. This is because the unit step function is only defined for non-negative values of t, and the exponential function is defined for all real numbers.
 

1. What is the significance of the limit of u(t) - u(t - δ) as δ goes to zero in LTI systems?

The limit of u(t) - u(t - δ) as δ goes to zero is an important concept in LTI (linear time-invariant) systems. It represents the step response of the system, which is the output when the input is a unit step function. This response helps in understanding the behavior of the system and can be used to analyze its stability and other characteristics.

2. How is the limit of u(t) - u(t - δ) as δ goes to zero related to the impulse response of an LTI system?

In an LTI system, the impulse response is defined as the output when the input is an impulse function. This response is closely related to the limit of u(t) - u(t - δ) as δ goes to zero. In fact, the impulse response can be obtained by taking the derivative of the step response. This is because the impulse function is the derivative of the unit step function.

3. Can the limit of u(t) - u(t - δ) as δ goes to zero be used to determine the stability of an LTI system?

Yes, the limit of u(t) - u(t - δ) as δ goes to zero can be used to determine the stability of an LTI system. If the limit exists and is finite, the system is stable. If the limit does not exist or is infinite, the system is unstable. This is because the step response gives information about the behavior of the system over time, and an unstable system will have a step response that grows without bound.

4. Is the limit of u(t) - u(t - δ) as δ goes to zero affected by the initial conditions of an LTI system?

No, the limit of u(t) - u(t - δ) as δ goes to zero is not affected by the initial conditions of an LTI system. This is because the unit step function is a causal input, meaning it starts at t=0 and has no effect for negative values of t. Therefore, the initial conditions do not play a role in determining the limit of the step response.

5. How is the limit of u(t) - u(t - δ) as δ goes to zero related to the transfer function of an LTI system?

The limit of u(t) - u(t - δ) as δ goes to zero is related to the transfer function of an LTI system in a few ways. First, the transfer function can be obtained from the step response by taking the Laplace transform. Additionally, the poles of the transfer function can be used to determine the stability of the system, which is also reflected in the limit of the step response. Furthermore, the limit of u(t) - u(t - δ) as δ goes to zero can be used to determine the frequency response of the system, which is another important characteristic described by the transfer function.

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