Continuous time signals - Causal signal given even part

In summary, to find the function h(t) for all t, you can use the definition of an even function and substitute -t for t in the given equation for h_e(t) for t<0. This results in the final function h(t) being equal to t[u(t)-u(t-1)]+u(t-1) for t>0 and -t[u(-t)-u(-t-1)]+u(-t-1) for t<0.
  • #1
lveenis
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Homework Statement


Suppose h(t) is a causal signal and has the even part h_e(t) given by:
h_e(t)= t[u(t)-u(t-1)]+u(t-1) for t>0.

Find h(t) for all t

Homework Equations


For an even function f(x), f(x) = f(-x)
Also even functions can be expressed as x_e(t) = 1/2[x(t)+x(-t)]

The Attempt at a Solution


I'm unsure how to approach this problem.
My assumption is that you have to find the odd part of the function and then sum the odd and the even components, since any function can be written as x(t)=x_e(t)+x_o(t) where x_e and x_o are even and odd functions.

So far I've gotten that the plot of the even part looks like this:
linear line with slope=1 from t=0-1 and then a straight line at 1 for t>1.
EDIT: not 0 for t<0 since this is defined for t>0 and it's even so it must be symmetric about the y-axis?

Since it says that this part is even and we know that that means h_e(t)=h_e(-t), does this means that it's symmetric about the y-axis and the question is asking me to find the odd function that essentially cancels out the negative part (making it causal) and thus amplifies the positive part?

My biggest problem is understanding exactly what the question is asking me to do; if anyone could tell me if my attempts are right or how to proceed I would appreciate it! Thank you!
 
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  • #2


Hello,

Yes, your understanding of the even and odd components of a function is correct. And yes, since the given function h_e(t) is defined only for t>0 and is even, it must be symmetric about the y-axis.

To find the function h(t) for all t, you can simply use the definition of an even function: h(t) = h_e(t) for t>0 and h(t) = h_e(-t) for t<0. This means that for t>0, h(t) will be the same as h_e(t) which you have already found. And for t<0, h(t) will be the same as h_e(-t), which can be found by simply substituting -t for t in the given equation for h_e(t).

Thus, the final function h(t) can be written as:

h(t) = t[u(t)-u(t-1)]+u(t-1) for t>0
h(t) = -t[u(-t)-u(-t-1)]+u(-t-1) for t<0

I hope this helps! Let me know if you have any further questions.
 

What is a continuous time signal?

A continuous time signal is a type of signal that exists in a continuous domain, meaning it has an infinite number of values within a given time interval. This is in contrast to a discrete time signal, which only has values at specific time points.

What does it mean for a signal to be causal?

A causal signal is one in which the output at any given time is only dependent on past or current inputs. This means that the signal does not have any future knowledge or dependence on future inputs.

What is the even part of a causal signal?

The even part of a causal signal is the portion of the signal that is symmetric about the y-axis. This means that for every value x, the signal also has a value at -x. In other words, the even part of a signal is its mirror image about the y-axis.

How is the even part of a causal signal calculated?

The even part of a causal signal can be calculated by taking the average of the original signal and its time-reversed version. This will result in a new signal that is symmetric about the y-axis, representing the even part of the original signal.

What is the importance of the even part in signal processing?

The even part of a signal is important in signal processing because it represents the symmetric component of the signal. This can be useful in filtering out unwanted noise or in understanding the overall shape and characteristics of the original signal.

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