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timofeeve [1]
3 years ago
5

Linear Time Invariant Systems For each of the systems below an input x(t) and the output y(t) are plotted. Determine whether eac

h of these are linear time invariant (LTI) systems, and can be described by a convolution. Provide an argument for your answer. Note that some cases are described in the time domain, and some in the frequency domain.
Engineering
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

The system can be described by a convolution

Explanation:

Thinking process:

If we consider a discrete input to a linear time-invariant system, then the system will be periodic with respect to the period, say N. This therefore, means that the output must also  be periodic. The proof is as follows:

The LTI system can be written for the system where:

y (n+N) = ∑h(k)x(n + N - k)

            = ∑h(k)x(n-k)\\= y(n)

From the proof, it turns out that y(y + N) = y(n) for any value of n, then the output will be the periodic with the period N.

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Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre
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Answer: \dot m_{in} = 23.942 \frac{kg}{s}, \dot H_{out} = 39632.62 kW

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to \dot V = 1 \frac{m^{3}}{s}. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

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- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0

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-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0

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Inflow (Superheated Steam)

\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}

The mass flow rate can be calculated by using this expression:

\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}

\dot m_{in} = 23.942 \frac{kg}{s}

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h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}

h_{out} = 1655.36 \frac{kJ}{kg}

The enthalpy rate at outflow is:

\dot H_{out} = \dot m \cdot h_{out}

\dot H_{out} = 39632.62 kW

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