Media Summary: If you the clock a bar H of Omega magnitude and then this is omega over omega squared plus No so we have i think 60 years and it's not section wise it's not such a nice okay anita is What a paul is okay a pole is a point where the function like the laplace transform goes infinity so s equals to

Uiuc Ece 210 18 1 - Detailed Analysis & Overview

If you the clock a bar H of Omega magnitude and then this is omega over omega squared plus No so we have i think 60 years and it's not section wise it's not such a nice okay anita is What a paul is okay a pole is a point where the function like the laplace transform goes infinity so s equals to Monochromatic by its name suggested that there's only So summing that gives me b minus so i have v minus being equal to v 0 ... the two distinguishing features okay so if these are the phasor input phasor output for frequency

Okay that we know how to do right then you get e to the j alpha t / j alpha 0 to t and then you get e to the j alpha t minus

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UIUC ECE 210 18-1
UIUC ECE 210 18-2
UIUC ECE 210 24-1
UIUC ECE 210 L49-1
UIUC ECE 210 1-1
UIUC ECE 210 L48-1
UIUC ECE 210 19-1
UIUC ECE 210 L51-1
UIUC ECE 210 11-1
UIUC ECE 210 1-2
UIUC ECE 210 20-1
UIUC ECE 210 24-2
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UIUC ECE 210 18-1

UIUC ECE 210 18-1

No no no

UIUC ECE 210 18-2

UIUC ECE 210 18-2

Sixth

UIUC ECE 210 24-1

UIUC ECE 210 24-1

If you the clock a bar H of Omega magnitude and then this is omega over omega squared plus

UIUC ECE 210 L49-1

UIUC ECE 210 L49-1

Minus

UIUC ECE 210 1-1

UIUC ECE 210 1-1

No so we have i think 60 years and it's not section wise it's not such a nice okay anita is

UIUC ECE 210 L48-1

UIUC ECE 210 L48-1

What a paul is okay a pole is a point where the function like the laplace transform goes infinity so s equals to

UIUC ECE 210 19-1

UIUC ECE 210 19-1

Monochromatic by its name suggested that there's only

UIUC ECE 210 L51-1

UIUC ECE 210 L51-1

Okay and in this example everything is

UIUC ECE 210 11-1

UIUC ECE 210 11-1

So summing that gives me b minus so i have v minus being equal to v 0

UIUC ECE 210 1-2

UIUC ECE 210 1-2

Charge 10 to the minus 90. so

UIUC ECE 210 20-1

UIUC ECE 210 20-1

1

UIUC ECE 210 24-2

UIUC ECE 210 24-2

... the two distinguishing features okay so if these are the phasor input phasor output for frequency

UIUC ECE 210 L28-1

UIUC ECE 210 L28-1

Okay that we know how to do right then you get e to the j alpha t / j alpha 0 to t and then you get e to the j alpha t minus