Bode Plot: Example 7

Draw the Bode Diagram for the transfer function:

This is the same as "Example 1," but has a 0.01 second time delay.  We have not seen a time delay before this, but we can easily handle it as we would any other constituent part of the transfer function. The magnitude and phase of a time delay are described here.

Step 1: Rewrite the transfer function in proper form.  

Make both the lowest order term in the numerator and denominator unity.  The numerator is an order 0 polynomial, the denominator is order 1.

Step 2: Separate the transfer function into its constituent parts.

The transfer function has 3 components:

• A constant of 3.3
• A pole at s=-30
• A time delay of 0.01 seconds (magnitude and phase of time delay described here).

Step 3: Draw the Bode diagram for each part.

This is done in the diagram below.

• The constant is the cyan line (A quantity of 3.3 is equal to 10.4 dB).  The phase is constant at 0 degrees.
• The pole at 30 rad/sec is the blue line.  It is 0 dB up to the break frequency, then drops off with a slope of -20 dB/dec.  The phase is 0 degrees up to 1/10 the break frequency (3 rad/sec) then drops linearly down to -90 degrees at 10 times the break frequency (300 rad/sec).
• The time delay is the red line.  It is 0 dB at all frequencies.  The phase  of the time delayis given by -0.01·ω rad, or -0.01·ω·180/π° (at ω=100 rad/sec, the phase is -0.01·100·180/π≈-30°).  There is no asymptotic approximation for the phase of a time delay.  Though the equation for the phase is linear with frequency, it looks exponential on the graph because the horizontal axis is logarithmic.

Step 4:  Draw the overall Bode diagram by adding up the results from step 3.

The exact response is the black line.