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Tutorial 8A (Complex Numbers) Complex numbers Q1) Evaluate $z=(2-i)^{2}+\frac{7...
Oct 29, 2024
Solution by Steps
For Q1:
step 1
To evaluate z=(2−i)2+7−4i2+i−11+i z = (2-i)^{2} + \frac{7-4i}{2+i} - \frac{1}{1+i} , we first calculate (2−i)2 (2-i)^{2} :
step 2
(2−i)2=4−4i+(−1)=3−4i (2-i)^{2} = 4 - 4i + (-1) = 3 - 4i
step 3
Next, we simplify 7−4i2+i \frac{7-4i}{2+i} by multiplying the numerator and denominator by the conjugate 2−i 2-i :
step 4
(7−4i)(2−i)(2+i)(2−i)=14−7i−8i+44+1=18−15i5=185−3i \frac{(7-4i)(2-i)}{(2+i)(2-i)} = \frac{14 - 7i - 8i + 4}{4 + 1} = \frac{18 - 15i}{5} = \frac{18}{5} - 3i
step 5
Now, simplify −11+i -\frac{1}{1+i} by multiplying by the conjugate 1−i 1-i :
step 6
−1(1−i)(1+i)(1−i)=−1−i2=−12+12i -\frac{1(1-i)}{(1+i)(1-i)} = -\frac{1-i}{2} = -\frac{1}{2} + \frac{1}{2}i
step 7
Combine all parts: z=(3−4i)+(185−3i)+(−12+12i) z = (3 - 4i) + \left(\frac{18}{5} - 3i\right) + \left(-\frac{1}{2} + \frac{1}{2}i\right)
step 8
Combine real and imaginary parts: z=(3+185−12)+(−4−3+12)i z = \left(3 + \frac{18}{5} - \frac{1}{2}\right) + \left(-4 - 3 + \frac{1}{2}\right)i
step 9
Calculate the real part: 3+185−12=3010+3610−510=6110 3 + \frac{18}{5} - \frac{1}{2} = \frac{30}{10} + \frac{36}{10} - \frac{5}{10} = \frac{61}{10}
step 10
Calculate the imaginary part: −4−3+12=−7+12=−142+12=−132 -4 - 3 + \frac{1}{2} = -7 + \frac{1}{2} = -\frac{14}{2} + \frac{1}{2} = -\frac{13}{2}
step 11
Thus, z=6110−132i z = \frac{61}{10} - \frac{13}{2}i in rectangular form
step 12
To convert to polar form, find r=(6110)2+(−132)2 r = \sqrt{\left(\frac{61}{10}\right)^{2} + \left(-\frac{13}{2}\right)^{2}} and θ=tan⁡−1(−1326110) \theta = \tan^{-1}\left(\frac{-\frac{13}{2}}{\frac{61}{10}}\right)
step 13
Finally, express in complex exponential form as z=rejθ z = re^{j\theta}
Answer
z=6110−132i z = \frac{61}{10} - \frac{13}{2}i in rectangular form, polar form r r and θ \theta calculated, and complex exponential form rejθ re^{j\theta}
Key Concept
Complex numbers can be expressed in rectangular, polar, and exponential forms.
Explanation
The evaluation of z z involves simplifying each term and combining them, leading to different forms of representation.
--- For Q2:a)
step 1
For the quadratic y=−2x2+3x−c y = -2x^{2} + 3x - c to have one repeated root, the discriminant must be zero: b2−4ac=0 b^{2} - 4ac = 0
step 2
Here, a=−2,b=3,c=−c a = -2, b = 3, c = -c , so 32−4(−2)(−c)=0 3^{2} - 4(-2)(-c) = 0 gives 9−8c=0 9 - 8c = 0
step 3
Solving for c c : 8c=9 8c = 9 leads to c=98 c = \frac{9}{8}
b)
step 4
For c=−2 c = -2 , the equation becomes y=−2x2+3x+2 y = -2x^{2} + 3x + 2
step 5
The roots are found using the quadratic formula: x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
step 6
Here, a=−2,b=3,c=2 a = -2, b = 3, c = 2 : x=−3±32−4(−2)(2)2(−2)=−3±9+16−4=−3±5−4 x = \frac{-3 \pm \sqrt{3^{2} - 4(-2)(2)}}{2(-2)} = \frac{-3 \pm \sqrt{9 + 16}}{-4} = \frac{-3 \pm 5}{-4}
step 7
This gives roots x=2−4=−12 x = \frac{2}{-4} = -\frac{1}{2} and x=−8−4=2 x = \frac{-8}{-4} = 2
c)
step 8
As c→+∞ c \rightarrow +\infty , the parabola opens downwards and will have no real roots (complex roots)
d)
step 9
As c→−∞ c \rightarrow -\infty , the parabola opens downwards and will have two real roots
Answer
a) c=98 c = \frac{9}{8} , b) Roots are −12 -\frac{1}{2} and 2 2 , c) Complex roots, d) Two real roots
Key Concept
The discriminant of a quadratic determines the nature of its roots.
Explanation
A zero discriminant indicates a repeated root, while positive and negative values indicate real and complex roots, respectively.
--- For Q3:
step 1
Given z=(1+2i)‾=1−2i z = \overline{(1 + 2i)} = 1 - 2i , we need to evaluate z4 z^{4}
step 2
First, calculate z2=(1−2i)2=1−4i+4(−1)=−3−4i z^{2} = (1 - 2i)^{2} = 1 - 4i + 4(-1) = -3 - 4i
step 3
Now, calculate z4=(−3−4i)2=9+24i+16(−1)=−7+24i z^{4} = (-3 - 4i)^{2} = 9 + 24i + 16(-1) = -7 + 24i
Answer
z4=−7+24i z^{4} = -7 + 24i
Key Concept
The conjugate of a complex number can be raised to a power using binomial expansion.
Explanation
The evaluation of z4 z^{4} involves squaring the complex number twice, leading to the final result.
--- For Q4:
step 1
Given (a−bi)2(3+5i)=(−6−24i)‾ (a - bi)^{2}(3 + 5i) = \overline{(-6 - 24i)} , we first find (−6−24i)‾=−6+24i \overline{(-6 - 24i)} = -6 + 24i
step 2
Let (a−bi)2=x (a - bi)^{2} = x , then x(3+5i)=−6+24i x(3 + 5i) = -6 + 24i
step 3
Solving for x x : x=−6+24i3+5i x = \frac{-6 + 24i}{3 + 5i}
step 4
Multiply by the conjugate: x=(−6+24i)(3−5i)(3+5i)(3−5i)=−18+30i+72i+1209+25=102+102i34=3+3i x = \frac{(-6 + 24i)(3 - 5i)}{(3 + 5i)(3 - 5i)} = \frac{-18 + 30i + 72i + 120}{9 + 25} = \frac{102 + 102i}{34} = 3 + 3i
step 5
Thus, (a−bi)2=3+3i (a - bi)^{2} = 3 + 3i
step 6
Taking the square root gives a−bi=3+3i a - bi = \sqrt{3 + 3i}
step 7
To find a a and b b , express 3+3i \sqrt{3 + 3i} in polar form and then convert back to rectangular form
Answer
a a and b b can be found from a−bi=3+3i a - bi = \sqrt{3 + 3i} after converting to polar form.
Key Concept
The square root of a complex number can be found using polar coordinates.
Explanation
The solution involves manipulating the equation and using polar representation to find the values of a a and b b .
--- For Q5:a)
step 1
Given x=[2,0,0,3]T x = [2, 0, 0, 3]^{T} and N=4 N = 4 , the DFT coefficients X X are calculated using the formula:
step 2
X[k]=∑n=0N−1x[n]e−j2πNkn X[k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{2\pi}{N} kn} for k=0,1,2,3 k = 0, 1, 2, 3
step 3
For k=0 k = 0 : X[0]=2+0+0+3=5 X[0] = 2 + 0 + 0 + 3 = 5
step 4
For k=1 k = 1 : X[1]=2+0+0+3e−j2π4⋅3=2+3(−j)=2−3j X[1] = 2 + 0 + 0 + 3e^{-j \frac{2\pi}{4} \cdot 3} = 2 + 3(-j) = 2 - 3j
step 5
For k=2 k = 2 : X[2]=2+0+0+3e−j2π4⋅6=2+3(−1)=−1 X[2] = 2 + 0 + 0 + 3e^{-j \frac{2\pi}{4} \cdot 6} = 2 + 3(-1) = -1
step 6
For k=3 k = 3 : X[3]=2+0+0+3e−j2π4⋅9=2+3j X[3] = 2 + 0 + 0 + 3e^{-j \frac{2\pi}{4} \cdot 9} = 2 + 3j
step 7
Thus, X=[5,2−3j,−1,2+3j]T X = [5, 2 - 3j, -1, 2 + 3j]^{T}
b)
step 8
Given X=N⋅[0,3e−jπ4,0,0]T X = N \cdot [0, 3e^{-j \frac{\pi}{4}}, 0, 0]^{T} , we find x x via IDFT:
step 9
x[n]=1N∑k=0N−1X[k]ej2πNkn x[n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{j \frac{2\pi}{N} kn}
step 10
Here, N=4 N = 4 , so x[n]=14(0+3ej2π4n+0+0)=34ejπ2n x[n] = \frac{1}{4} \left(0 + 3e^{j \frac{2\pi}{4} n} + 0 + 0\right) = \frac{3}{4} e^{j \frac{\pi}{2} n}
step 11
Thus, x[n]=34sin⁡(π2n) x[n] = \frac{3}{4} \sin\left(\frac{\pi}{2} n\right)
c)
step 12
Given X=N⋅[2,3e−jπ4,0,0]T X = N \cdot [2, 3e^{-j \frac{\pi}{4}}, 0, 0]^{T} , we find x x similarly:
step 13
x[n]=14(2+3ej2π4n)=12+34ejπ2n x[n] = \frac{1}{4} \left(2 + 3e^{j \frac{2\pi}{4} n}\right) = \frac{1}{2} + \frac{3}{4} e^{j \frac{\pi}{2} n}
step 14
Thus, x[n]=12+34sin⁡(π2n) x[n] = \frac{1}{2} + \frac{3}{4} \sin\left(\frac{\pi}{2} n\right)
Answer
a) X=[5,2−3j,−1,2+3j]T X = [5, 2 - 3j, -1, 2 + 3j]^{T} , b) x[n]=34sin⁡(π2n) x[n] = \frac{3}{4} \sin\left(\frac{\pi}{2} n\right) , c) x[n]=12+34sin⁡(π2n) x[n] = \frac{1}{2} + \frac{3}{4} \sin\left(\frac{\pi}{2} n\right)
Key Concept
The Discrete Fourier Transform (DFT) converts time-domain signals into frequency-domain representations.
Explanation
The DFT coefficients are calculated from the time-domain signal, and the Inverse DFT (IDFT) reconstructs the time-domain signal from the frequency-domain coefficients.
--- For Q6:a)
step 1
The DFT Analysis matrix W W for N=16 N = 16 is defined as Wnk=e−j2πNnk W_{nk} = e^{-j \frac{2\pi}{N} nk}
step 2
Thus, W=[1amp;1amp;1amp;…1amp;e−j2π16amp;e−j4π16amp;…⋮amp;⋮amp;⋮amp;⋱] W = \begin{bmatrix} 1 & 1 & 1 & \ldots \\ 1 & e^{-j \frac{2\pi}{16}} & e^{-j \frac{4\pi}{16}} & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}
b)
step 3
The equation representing the kth k^{th} row of W W is Wk=[e−j2π16kn]n=015 W_k = \left[ e^{-j \frac{2\pi}{16} kn} \right]_{n=0}^{15}
step 4
The first 4 terms are [1,e−j2π16k,e−j4π16k,e−j6π16k] [1, e^{-j \frac{2\pi}{16} k}, e^{-j \frac{4\pi}{16} k}, e^{-j \frac{6\pi}{16} k}]
c)
step 5
The kth k^{th} row of W W relates to the DFT coefficient X(k) X(k) as it represents the basis functions used to compute X(k) X(k)
step 6
Each row corresponds to a specific frequency component in the DFT
d)
step 7
Given x(n)=3ej(2π16n+π3) x(n) = 3 e^{j\left(\frac{2\pi}{16} n + \frac{\pi}{3}\right)} , we find X(k) X(k) using the DFT formula
step 8
The DFT will yield non-zero coefficients at specific frequencies corresponding to the exponential term
e)
step 9
Given x(n)=3sin⁡(2π16n+π3) x(n) = 3 \sin\left(\frac{2\pi}{16} n + \frac{\pi}{3}\right) , we find X(k) X(k) similarly
step 10
The sine function can be expressed in terms of complex exponentials, leading to different coefficients compared to the previous case
Answer
a) DFT matrix W W defined, b) kth k^{th} row equation and first 4 terms listed, c) Relation to X(k) X(k) explained, d) X(k) X(k) for x(n)=3ej(2π16n+π3) x(n) = 3 e^{j\left(\frac{2\pi}{16} n + \frac{\pi}{3}\right)} , e) X(k) X(k) for sine function explained
Key Concept
The DFT matrix and its rows represent the transformation of time-domain signals into frequency-domain coefficients.
Explanation
Each row of the DFT matrix corresponds to a frequency component, and the transformation of different types of signals (exponential vs. sine) yields different frequency-domain representations.
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