State-Space System Analysis MCQ Quiz – Objective Question with Answer for State-Space System Analysis

Although the state space or the internal description of the system still involves a relationship between the input and output signals, it also involves an additional set of variables, called State variables.

The state variables provide information about all the internal signals in the system. As a result, the state-space description provides a more detailed description of the system than the input-output description.

We define the state of a system at time n0 as the amount of information that must be provided at time n0, which, together with the input signal x(n) for n≥n0 determines the output signal for n≥n0.

According to the definition of the state of a system, the system consists of two components called the memory component and memory less component.

The transpose of the matrix F is obtained by interchanging its rows and columns, and it is denoted by FT. The system thus obtained is known as Transposed system.

If h(n) is the impulse response of the single input single output system, and h1(n) is the impulse response of the transposed system, then we know that h(n)=h1>(n). Thus, a single input single output system and its transpose have identical impulse responses and hence the same input-output relationship.

A closed-form solution of the state space equations is easily obtained when the system matrix F is diagonal. Hence, by finding a matrix P so that F1=PFP-1 is diagonal, the solution of the state equations is simplified considerably.

A number λ is an Eigenvalue of F and a nonzero vector U is the associated Eigenvector if
FU=λU
Thus, we obtain (F-λI)U=0.

We know that (F-λI)U=0
The above equation has a nonzero solution U if the matrix F-λI is singular, which is the case if the determinant of (F-λI) is zero. That is, |F-λI|=0.

This determinant yields the characteristic polynomial of the matrix F.

The parallel form realization is also known as normal form representation, because the matrix F is diagonal, and hence the state variables are uncoupled.

(101.01)2=1*22+0*21+1*20+0*2-1+1*2-2=(5.25)10
=>x=5.25.

A real number X can be represented as X=\(\sum_{i=-A}^B b_i r^{-i}\) where bi represents the digit, ‘r’ is the radix or base, A is the number of integer digits, and B is the number of fractional digits.

The binary point between the digits b0 and b1 does not exist physically in the computer. Simply, the logic circuits of the computer are designed such that the computations result in numbers that correspond to the assumed location of this point.

A fixed point representation of numbers allows us to cover a range of numbers, say, x_max-x_min with a resolution

Δ=(x_max-x_min)/(m-1)

where m=2b is the number of levels and ‘b’ is the number of bits.

We can represent 5 as
5=0.625*8=0.625*23
The above number can be represented in binary float point representation as 0.101000*2011
Thus Mantissa=0.101000, Exponent=011.

Let us consider two numbers X=M.2E and Y=N.2F
If we multiply both X and Y, we get X.Y=(M.N).2E+F
Thus if we multiply two numbers, the mantissa is multiplied and the exponents are added.

Let the mantissa be represented by 23 bits plus a sign bit and let the exponent be represented by 7 bits plus a sign bit.

Thus, the smallest floating-point number that can be represented using the 32-bit number is
(1/2)*2^-127 = 0.3*10^-38

Thus, the smallest floating-point number that can be represented using the 32-bit number is
(1-2-23)*2¹²⁷ = 1.7*10³⁸.

According to the IEEE 754 standard, for a 32-bit machine, a single-precision floating-point number is represented as X=(-1)^s.2^E-127(M).
From the above equation, we can interpret that,
If 0<E<255, then X=(-1)^s.2^E-127(1.M)=>X is a mixed number.

For a two’s complement representation, the truncation error is always negative and falls in the range
-(2-b-2-bm) ≤ Et ≤ 0.

In floating-point representation, the mantissa is either rounded or truncated. Due to non-uniform resolution, the corresponding error in a floating-point representation is proportional to the number being quantized.

The number (-3/8) is stored in the computer as the 2’s complement of (3/8)
We know that the binary equivalent of (3/8)=0011
Thus the two’s complement of 0011=1101.

The binary floating-point representation commonly used in practice consists of a mantissa M, which is the fractional part of the number and falls in the range 1/2<M<1, multiplied by the exponential factor 2E, where the exponent E is either a negative or positive integer. Hence a number X is represented as X= M.2^E(1/2<M<1).

We can represent the numbers in binary float point as
5=0.101000(2⁰¹¹)
3/8=0.110000(2¹⁰¹)=0.000011(2⁰¹¹)
=>5+3/8=(0.101000+0.000011)(2⁰¹¹)=(0.101011)(2⁰¹¹)
Therefore mantissa=0.101011 and exponent=011.