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I was reading this PDF which tries to explain the concept of poles and zeros of Laplace transform.

My question is about FIGURE 32-5(b) on page #591.

enter image description here

I don't understand why the area is said to be zero when, as shown, it should clearly be some negative number. The mentioned PDF does try to give an answer which I don't understand.

Figure (b) shows one of the special cases we have been looking for. When this waveform is multiplied by the impulse response, the resulting integral has a value of zero. This occurs because the area above the x-axis (from the delta function) is exactly equal to the area below (from the rectified sinusoid). The values for σ and ω that produce this type of cancellation are called a zero of the system. As shown in the s-plane diagram of Fig. 32-4, zeros are indicated by small circles.

You can consider it to be a follow-on question to this comment where @AJN tried to explain it further but unfortunately I still couldn't get it. I thought it'd be better to ask it separately. I'd appreciate if you could help me with it?

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The dirac-delta function has an area of 1 when integrated. The red area and the delta function have the same area. When integrated from t = (0-) to t -> inf the total area is 0.

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  • \$\begingroup\$ Thank you. Just to clarify. Do you mean that the red area (negative) and the area of dirac delta (positive) cancel each other? Just wondering that why they chose to show a non-zero area? Mistake? \$\endgroup\$ – PG1995 Jun 16 at 3:05
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    \$\begingroup\$ The sum of all positive and negative values is zero. It is the integral of the entire waveform, not an instance of ti=x. \$\endgroup\$ – user105652 Jun 16 at 3:05
  • \$\begingroup\$ @Sparky256 Thank you. Sorry, I edited my previous comment once you had replied. Would appreciate it if you could comment on it. "Just wondering that why they chose to show a non-zero area? Mistake?" \$\endgroup\$ – PG1995 Jun 16 at 3:18
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    \$\begingroup\$ @PG1995 It's not a mistake. Remember that the Dirac delta function at t=0 has finite (and positive) area--this exactly cancels out the negative area highlighted in red. \$\endgroup\$ – Hearth Jun 16 at 3:35
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    \$\begingroup\$ They didn't choose to show a non-zero area. They chose to accurately depict the negative component of the area; and since it's impossible to accurately depict the positive component of the area, they chose to denote it using an arrow symbol. \$\endgroup\$ – Tanner Swett Jun 16 at 15:44

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