Answer:
Yes, a minimum phase continuous time system is also minimum phase when converted into discrete time system using bilinear transformation.
Step-by-step explanation:
Bilinear Transform:
In digital signal processing, the bilinear transform is used to convert continuous time system into discrete time system representation.
Minimum-Phase:
We know that a system is considered to be minimum phase if the zeros are situated in the left half of the s-plane in continuous time system. In the same way, a system is minimum phase when its zeros are inside the unit circle of z-plane in discrete time system.
The bilinear transform is used to map the left half of the s-plane to the interior of the unit circle in the z-plane preserving the stability and minimum phase property of the system. Therefore, a minimum phase continuous time system is also minimum phase when converted into discrete time system using bilinear transformation.
Answer:
(6x+3)(6x-3)
Step-by-step explanation:
I used FOIL
Multiply the first numbers 6x*6x=36x^2
The outer -18x
Inner 18x which the outer and the inner cross out
Last 3*-3=-9
36x^2 -9
Answer:
The second answer is correct (5 x 1,000,000,000)
Step-by-step explanation:
Just count the number of zeros. Multiplying a number liket his by five doesn't change the number of zeros
Answer:
f(x) = 2x + 4 domains {-1, 0, 1}
range {2, 4, 6}
(please mark brain if this helps and is correct)
Step-by-step explanation:
to solve f(x) to find the range you would want to input all the domain numbers in the equation f(x) to get the range of all the numbers you need
step 1: take the first domain number and input it into your equation where x is
ex: 2x + 4 [2(-1) + 4] = -2 + 4 = 2
step 2: add the second domain number and input it into your equation where x is
ex: 2x + 4 [2(0) + 4] = 0 + 4 = 4
step 3: add the last domain number and input it into your equation where x is
ex: 2x + 4 [2(1) + 4] = 2 + 4 = 6
Now we have determined what the range is
Answer:
Step-by-step explanation:
Vertical Asymptote: x=2Horizontal Asymptote: NoneEquation of the Slant/Oblique Asymptote: y=x 3+23 Explanation:Given:y=f(x)=x2−93x−6Step.1:To find the Vertical Asymptote:a. Factor where possibleb. Cancel common factors, if anyc. Set Denominator = 0We will start following the steps:Consider:y=f(x)=x2−93x−6We will factor where possible:y=f(x)=(x+3)(x−3)3x−6If there are any common factors in the numerator and the denominator, we can cancel them.But, we do not have any.Hence, we will move on.Next, we set the denominator to zero.(3x−6)=0Add 6 to both sides.(3x−6+6)=0+6(3x−6+6)=0+6⇒3x=6⇒x=63=2Hence, our Vertical Asymptote is at x=2Refer to the graph below:enter image source hereStep.2:To find the Horizontal Asymptote:Consider:y=f(x)=x2−93x−6Since the highest degree of the numerator is greater than the highest degree of the denominator,Horizontal Asymptote DOES NOT EXISTStep.3:To find the Slant/Oblique Asymptote:Consider:y=f(x)=x2−93x−6Since, the highest degree of the numerator is one more than the highest degree of the denominator, we do have a Slant/Oblique AsymptoteWe will now perform the Polynomial Long Division usingy=f(x)=x2−93x−6enter image source hereHence, the Result of our Long Polynomial Division isx3+23+(−53x−6)
