How do you add a multiplication in the numerator of a transfer function.

The equation I need to represent as a transfer function is 5*e-ts/s+5. I coded as follows but my numerator returns as + 5 i need it to be * 5. Is there a way to achieve this?

clc
clear all
num=[exp(-1) 5];
den=[1 5];
g=tf(num,den)

g =

0.3679 s + 5

------------

s + 5

Answer 1

[Star Strider]

Perhaps this —

num=5*exp(-1);
den=[1 5];
g=tf(num,den)

g =
 
  1.839
  -----
  s + 5
 
Continuous-time transfer function.

The exponential function here is a constant. If you intend to introduce a delay, use InputDelay or one of the other options.

EDIT — (16 Apr 2023 at 10:44)

The transfer function would be —

num = 5;
den = [1 5];
g = tf(num, den, 'InputDelay',1)

g =
 
                5
  exp(-1*s) * -----
              s + 5
 
Continuous-time transfer function.

.

[Walter Roberson]

Side note: Using tf('s'), exp(-s) * expression has the same internal effect as using 'InputDelay', 1 . Using 'OutputDelay', 1 has very nearly the same effect but sets an internal OutputDelay instead of InputDelay, but displays the same. Later when you go to do calculations based on the tf created, the (up to) 2 delays are merged.

I would have to check to be sure, but my memory is telling me that for calculation purposes, the input delay and output delay get combined to input delay and the result cannot be told apart from the case where you had specified the combined delay as an input delay. I had always wondered why they bother to separate them out, as I never came across a case where the two kinds of delays had different events.

[Star Strider]

Therer is also IODelay (‘Transport Delay’ as I originally wrote) that may combine them. The documentation doesn’t appear to go into much detail on them.

[Paul]

@Walter Roberson

InputDelays and OutputDelays, in general, have to be treated separately for systems with multiple inputs and/or multiple outputs.

Also, InputDelay and OutputDelay affects the evolution of state variables in an ss model, even in the single-input, single-output case

h = ss(tf(1,[1 1]));
sys1 = h; set(sys1,'InputDelay',1);
sys2 = h; set(sys2,'OutputDelay',1);

[y1,t1,x1] = step(sys1);
[y2,t2,x2] = step(sys2);
figure
plot(t1,y1,t2,y2),grid

figure
plot(t1,x1,t2,x2),grid

[Star Strider]

I am not certain how to use τ as a unit step functon in this instance.

Perhaps this —

I = imread('matlab.jpg');
imshow(I)

num = 5;
den = [1 5];
g = tf(num, den, 'InputDelay',heaviside(1))

g =
 
                5
  exp(-1*s) * -----
              s + 5
 
Continuous-time transfer function.

The heaviside function has to have an argument, so I chose 1.

.

[Walter Roberson]

heaviside(1) is 1.

num = 5;
den = [1 5];
g = tf(num, den, 'InputDelay',1)

g =
 
                5
  exp(-1*s) * -----
              s + 5
 
Continuous-time transfer function.

[Star Strider]

Of course, however it is a function of time, with the time argument not otherwise defined. This seems to be getting a bit less certain.

Answer 2

[Walter Roberson]

s = tf('s');
g = 5 * exp(-t*s)/s + 5

Unrecognized function or variable 't'.

You cannot use an undefined or symbolic variable in a tf().

Are you sure that you want to add 5 to the delayed signal? Rather than using (s+5) as the denominator ?

https://www.mathworks.com/help/control/ug/time-delays-in-linear-systems.html

[Walter Roberson]

My guess is that you want

s = tf('s');
g = 5 * exp(-s)/(s + 5)

g =
 
                5
  exp(-1*s) * -----
              s + 5
 
Continuous-time transfer function.

[Sean Flanagan]

Sorry Walter and Star Trider the equation I tried to describe is the one attached to this as matlab.jpg. Tau is a unity step so I assumed it went to 0.

[Walter Roberson]

If you apply a unity step to s, the result would be the same as putting in the restriction that s >= 0

[Paul]

What does "Tau is a unity step" mean? Tau is place holder for a number, isn't it?

How can a unity step be applied to s? s is the independent variable of the Laplace transform. In that same context, what does "restriction that s>=0" mean?