I'm trying to graph a simple response function: 1/(1-0.5s^-1)Now, I know that the function can also be written as: s/(s-0.5)So I tried plotting the step and impulse responses in Matlab:sys = tf([1 0],[1 -0.5])figure(1);step(sys);figure(2);impulse(sys);However, both graphs look the same (can't post images of my graphs, I need more rep to do it).Both graphs have exponential growth, but shouldn't the impulse response look like exponential decay?

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I&#039;m trying to graph a simple response function: 1/(1-0.5s^-1)Now, I know that the function can also be written as: s/(s-0.5)So I tried plotting the step and impulse responses in Matlab:sys = tf([1 0],[1 -0.5])figure(1);step(sys);figure(2);impulse(sys);However, both graphs look the same (can&#039;t post images of my graphs, I need more rep to do it).Both graphs have exponential growth, but shouldn&#039;t the impulse response look like exponential decay?

Zackary Galloway · Accepted Answer

The problem is, you have a highly unstable transfer function. So you can&#039;t expect a decaying impulse/step-response. In other words:  H  (  s  )  =      s          s      −      0.5          ⟹    h  (  t  )  =            L              −      1        {  H  }  =  δ  (  t  )  +      1    2        e          t              /            2      For impulse-response:  y  (  t  )  =  h  (  t  )  ∗  δ  (  t  )  =  h  (  t  )and for step-response:  y  (  t  )  =      ∫    0    t    h  (  τ  )  d  τ  =  u  (  t  )  +      e          t              /            2      where ∗ stands for convolution. As you see, both responses have exponential growth in time.

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