# LTI System continuity

#### laguna92651

Joined Mar 29, 2008
101
I need to determine the value of alpha, α, for a Linear Time Invaraiant system to ensure continuity. 50e^(-αf) The waveform, H(f), is symmetrical

around 0, with a flat top at amplitude 3, then at frequency 150Hz the exponential, 50e^(-αf), starts. I'm not sure what is meant by ensuring continuity, so I don't know how to find α. And redraw showing the 3dB cutoff frequency.

#### steveb

Joined Jul 3, 2008
2,436
I need to determine the value of alpha, α, for a Linear Time Invaraiant system to ensure continuity. 50e^(-αf) The waveform, H(f), is symmetrical

around 0, with a flat top at amplitude 3, then at frequency 150Hz the exponential, 50e^(-αf), starts. I'm not sure what is meant by ensuring continuity, so I don't know how to find α. And redraw showing the 3dB cutoff frequency.
Continuity simply means that the values of the function must change without any sudden jumps. Here you are trying to match the boundary between two different functions. So, simply choose the value of alpha that allows the function to have the same value at the junction.

#### laguna92651

Joined Mar 29, 2008
101
Thanks Steve that makes sense. Does this calculation seem correct.

ln(e^(-
α150))=ln(3)
-α150ln(e)=ln(3)
-α150(1)=ln(3)
α=-1.0986/150
α=-0.0073

I assume the 3 is 3 dB, if that were the case the 3dB cutoff would be 0 on the y axis and the cutoff frequency would be out at infinity, would that be correct? What do you think about α= -0.0073?

#### steveb

Joined Jul 3, 2008
2,436
Thanks Steve that makes sense. Does this calculation seem correct.

ln(e^(-
α150))=ln(3)
-α150ln(e)=ln(3)
-α150(1)=ln(3)
α=-1.0986/150
α=-0.0073

I assume the 3 is 3 dB, if that were the case the 3dB cutoff would be 0 on the y axis and the cutoff frequency would be out at infinity, would that be correct? What do you think about α= -0.0073?
Hmmm, something doesn't seem right here. I agree with your calculation based on the functions in your graph, but a negative alpha is a problem. The negative of a negative alpha is a positive alpha. A positive alpha will make the exponential increase rather than decrease. Are you sure there isn't a scale factor on the exponential function? Above you seemed to have written 50*exp(-alpha*f).

On the other issue with the 3 dB cutoff, I'm not sure. I'll think about it a little more to make sure I agree because it could be a linear scale on the y-axis. But, first lets figure out what function you need to match.

#### laguna92651

Joined Mar 29, 2008
101
My fault is should be, 50*exp^(-alpha*f)

#### laguna92651

Joined Mar 29, 2008
101
I'm rusty on my natural log math, would it be ln(50)*ln(e)=3.91?

#### laguna92651

Joined Mar 29, 2008
101
Hi Steve
Fixed the sign problem.

50e^(-αf)

50e^(-α150) = 3
ln(50e^(-α150)) = ln(3)
ln(50) + ln(e^(-α150)) = ln(3)
3.91 -α150 = 1.0986
-α = (1.0986-3.91)/150
-α = -2.8114/150
α = .0187

50e^(-.0187f)
Would do you suggest for the the cutoff frequency?

#### steveb

Joined Jul 3, 2008
2,436
Hi Steve
Fixed the sign problem.

50e^(-αf)

50e^(-α150) = 3
ln(50e^(-α150)) = ln(3)
ln(50) + ln(e^(-α150)) = ln(3)
3.91 -α150 = 1.0986
-α = (1.0986-3.91)/150
-α = -2.8114/150
α = .0187

50e^(-.0187f)
Would do you suggest for the the cutoff frequency?
That looks correct to me. For the cutoff frequency, I would assume that the transfer function you have is on a linear scale, not on a dB scale. I can't be 100% sure about that, but technically the magnitude should be labeled 3 dB if that is what is meant. The fact that it is shown as 3 tells me it is a linear scale. The 3 dB down point is 0.707 of the peak value, on a linear scale.